1439 lines
40 KiB
TeX
1439 lines
40 KiB
TeX
\chapter{Range queries}
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\index{range query}
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\index{sum query}
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\index{minimum query}
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\index{maximum query}
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In this chapter, we discuss data structures
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that allow us to efficiently answer range queries.
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In a \key{range query}, we are given two indices
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to an array, and our task is to calculate some
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value based on the elements between the given indices.
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Typical range queries are:
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\begin{itemize}
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\item \key{sum query}: calculate the sum of elements
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\item \key{minimum query}: find the smallest element
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\item \key{maximum query}: find the largest element
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\end{itemize}
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For example, consider the range $[4,7]$ in the following array:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (3,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$8$};
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\node at (3.5,0.5) {$4$};
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\node at (4.5,0.5) {$6$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$3$};
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\node at (7.5,0.5) {$4$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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In this range, the sum of elements is $4+6+1+3=16$,
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the minimum element is 1 and the maximum element is 6.
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A simple way to process range queries is to
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go through all elements in the range.
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For example, the following function \texttt{sum}
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calculates the sum of elements in a range
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$[a,b]$ of an array $t$:
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\begin{lstlisting}
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int sum(int a, int b) {
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int s = 0;
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for (int i = a; i <= b; i++) {
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s += t[i];
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}
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return s;
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}
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\end{lstlisting}
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The above function works in $O(n)$ time,
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where $n$ is the number of elements in the array.
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Thus, we can process $q$ queries in $O(nq)$
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time using the function.
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However, if both $n$ and $q$ are large, this approach
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is slow, and it turns out that there are
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ways to process range queries much more efficiently.
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\section{Static array queries}
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We first focus on a situation where
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the array is \key{static}, i.e.,
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the elements are never modified between the queries.
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In this case, it suffices to construct
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a static data structure that tells us
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the answer for any possible query.
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\subsubsection{Sum queries}
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\index{prefix sum array}
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It turns out that we can easily process
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sum queries on a static array,
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because we can use a data structure called
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a \key{prefix sum array}.
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Each value in such an array equals
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the sum of values in the original array up to that position.
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For example, consider the following array:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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%\fill[color=lightgray] (3,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$4$};
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\node at (3.5,0.5) {$8$};
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\node at (4.5,0.5) {$6$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$4$};
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\node at (7.5,0.5) {$2$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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The corresponding prefix sum array is as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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%\fill[color=lightgray] (3,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$4$};
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\node at (2.5,0.5) {$8$};
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\node at (3.5,0.5) {$16$};
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\node at (4.5,0.5) {$22$};
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\node at (5.5,0.5) {$23$};
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\node at (6.5,0.5) {$27$};
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\node at (7.5,0.5) {$29$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Let $\textrm{sum}(a,b)$ denote the sum of elements
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in the range $[a,b]$.
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Since the prefix sum array contains all values
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of the form $\textrm{sum}(1,k)$,
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we can calculate any value of
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$\textrm{sum}(a,b)$ in $O(1)$ time, because
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\[ \textrm{sum}(a,b) = \textrm{sum}(1,b) - \textrm{sum}(1,a-1).\]
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By defining $\textrm{sum}(1,0)=0$,
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the above formula also holds when $a=1$.
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For example, consider the range $[4,7]$:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (3,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$4$};
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\node at (3.5,0.5) {$8$};
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\node at (4.5,0.5) {$6$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$4$};
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\node at (7.5,0.5) {$2$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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The sum in the range is $8+6+1+4=19$.
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This sum can be calculated using
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two values in the prefix sum array:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (2,0) rectangle (3,1);
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\fill[color=lightgray] (6,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$4$};
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\node at (2.5,0.5) {$8$};
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\node at (3.5,0.5) {$16$};
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\node at (4.5,0.5) {$22$};
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\node at (5.5,0.5) {$23$};
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\node at (6.5,0.5) {$27$};
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\node at (7.5,0.5) {$29$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Thus, the sum in the range $[4,7]$ is $27-8=19$.
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It is also possible to generalize this idea
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to higher dimensions.
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For example, we can construct a two-dimensional
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prefix sum array that can be used for calculating
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the sum of any rectangular subarray in $O(1)$ time.
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Each value in such an array is the sum of a subarray
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that begins at the upper-left corner of the array.
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\begin{samepage}
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The following picture illustrates the idea:
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\begin{center}
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\begin{tikzpicture}[scale=0.54]
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\draw[fill=lightgray] (3,2) rectangle (7,5);
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\draw (0,0) grid (10,7);
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%\draw[line width=2pt] (3,2) rectangle (7,5);
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\node[anchor=center] at (6.5, 2.5) {$A$};
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\node[anchor=center] at (2.5, 2.5) {$B$};
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\node[anchor=center] at (6.5, 5.5) {$C$};
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\node[anchor=center] at (2.5, 5.5) {$D$};
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\end{tikzpicture}
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\end{center}
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\end{samepage}
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The sum of the gray subarray can be calculated
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using the formula
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\[S(A) - S(B) - S(C) + S(D),\]
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where $S(X)$ denotes the sum of a rectangular
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subarray from the upper-left corner
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to the position of $X$.
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\subsubsection{Minimum queries}
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Next we will see how we can
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process range minimum queries in $O(1)$ time
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after an $O(n \log n)$ time preprocessing\footnote{There are also
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sophisticated techniques \cite{fis06} where the preprocessing time
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is only $O(n)$, but such algorithms are not needed in
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competitive programming.}.
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Note that minimum and maximum queries can always
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be processed using similar techniques,
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so it suffices to focus on minimum queries.
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Let $\textrm{rmq}(a,b)$ (''range minimum query'')
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denote the minimum element in the range $[a,b]$.
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The idea is to precalculate all values of $\textrm{rmq}(a,b)$
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where $b-a+1$, the length of the range, is a power of two.
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For example, for the array
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$4$};
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\node at (3.5,0.5) {$8$};
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\node at (4.5,0.5) {$6$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$4$};
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\node at (7.5,0.5) {$2$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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the following values will be calculated:
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\begin{center}
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\begin{tabular}{ccc}
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\begin{tabular}{ccc}
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$a$ & $b$ & $\textrm{rmq}(a,b)$ \\
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\hline
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1 & 1 & 1 \\
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2 & 2 & 3 \\
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3 & 3 & 4 \\
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4 & 4 & 8 \\
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5 & 5 & 6 \\
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6 & 6 & 1 \\
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7 & 7 & 4 \\
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8 & 8 & 2 \\
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\end{tabular}
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&
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\begin{tabular}{ccc}
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$a$ & $b$ & $\textrm{rmq}(a,b)$ \\
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\hline
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1 & 2 & 1 \\
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2 & 3 & 3 \\
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3 & 4 & 4 \\
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4 & 5 & 6 \\
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5 & 6 & 1 \\
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6 & 7 & 1 \\
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7 & 8 & 2 \\
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\\
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\end{tabular}
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&
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\begin{tabular}{ccc}
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$a$ & $b$ & $\textrm{rmq}(a,b)$ \\
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\hline
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1 & 4 & 1 \\
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2 & 5 & 3 \\
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3 & 6 & 1 \\
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4 & 7 & 1 \\
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5 & 8 & 1 \\
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1 & 8 & 1 \\
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\\
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\\
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\end{tabular}
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\end{tabular}
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\end{center}
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The number of precalculated values is $O(n \log n)$,
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because there are $O(\log n)$ range lengths
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that are powers of two.
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In addition, the values can be calculated efficiently
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using the recursive formula
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\[\textrm{rmq}(a,b) = \min(\textrm{rmq}(a,a+w-1),\textrm{rmq}(a+w,b)),\]
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where $b-a+1$ is a power of two and $w=(b-a+1)/2$.
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Calculating all those values takes $O(n \log n)$ time.
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After this, any value of $\textrm{rmq}(a,b)$ can be calculated
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in $O(1)$ time as a minimum of two precalculated values.
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Let $k$ be the largest power of two that does not exceed $b-a+1$.
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We can calculate the value of $\textrm{rmq}(a,b)$ using the formula
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\[\textrm{rmq}(a,b) = \min(\textrm{rmq}(a,a+k-1),\textrm{rmq}(b-k+1,b)).\]
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In the above formula, the range $[a,b]$ is represented
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as the union of the ranges $[a,a+k-1]$ and $[b-k+1,b]$, both of length $k$.
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As an example, consider the range $[2,7]$:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (1,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$4$};
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\node at (3.5,0.5) {$8$};
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\node at (4.5,0.5) {$6$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$4$};
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\node at (7.5,0.5) {$2$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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The length of the range is 6,
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and the largest power of two that does
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not exceed 6 is 4.
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Thus the range $[2,7]$ is
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the union of the ranges $[2,5]$ and $[4,7]$:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (1,0) rectangle (5,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$4$};
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\node at (3.5,0.5) {$8$};
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\node at (4.5,0.5) {$6$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$4$};
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\node at (7.5,0.5) {$2$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (3,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$1$};
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\node at (1.5,0.5) {$3$};
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\node at (2.5,0.5) {$4$};
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\node at (3.5,0.5) {$8$};
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\node at (4.5,0.5) {$6$};
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\node at (5.5,0.5) {$1$};
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\node at (6.5,0.5) {$4$};
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\node at (7.5,0.5) {$2$};
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|
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Since $\textrm{rmq}(2,5)=3$ and $\textrm{rmq}(4,7)=1$,
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we can conclude that $\textrm{rmq}(2,7)=1$.
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\section{Binary indexed trees}
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\index{binary indexed tree}
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\index{Fenwick tree}
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|
|
A \key{binary indexed tree} or \key{Fenwick tree}\footnote{The
|
|
binary indexed tree structure was presented by P. M. Fenwick in 1994 \cite{fen94}.}
|
|
can be seen as a dynamic version of a prefix sum array.
|
|
This data structure supports two $O(\log n)$ time operations:
|
|
calculating the sum of elements in a range
|
|
and modifying the value of an element.
|
|
|
|
The advantage of a binary indexed tree is
|
|
that it allows us to efficiently update
|
|
the array elements between the sum queries.
|
|
This would not be possible using a prefix sum array,
|
|
because after each update, we should build the
|
|
whole array again in $O(n)$ time.
|
|
|
|
\subsubsection{Structure}
|
|
|
|
A binary indexed tree can be represented as an array
|
|
where the value at position $x$
|
|
equals the sum of elements in the range $[x-k+1,x]$,
|
|
where $k$ is the largest power of two that divides $x$.
|
|
For example, if $x=6$, then $k=2$, because 2 divides 6
|
|
but 4 does not divide 6.
|
|
|
|
\begin{samepage}
|
|
For example, consider the following array:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$3$};
|
|
\node at (2.5,0.5) {$4$};
|
|
\node at (3.5,0.5) {$8$};
|
|
\node at (4.5,0.5) {$6$};
|
|
\node at (5.5,0.5) {$1$};
|
|
\node at (6.5,0.5) {$4$};
|
|
\node at (7.5,0.5) {$2$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
\begin{samepage}
|
|
The corresponding binary indexed tree is as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$4$};
|
|
\node at (2.5,0.5) {$4$};
|
|
\node at (3.5,0.5) {$16$};
|
|
\node at (4.5,0.5) {$6$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$4$};
|
|
\node at (7.5,0.5) {$29$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
For example, the value at position 6
|
|
in the binary indexed tree is 7,
|
|
because the sum of elements in the range $[5,6]$
|
|
of the array is $6+1=7$.
|
|
|
|
The following picture shows more clearly
|
|
how each value in the binary indexed tree
|
|
corresponds to a range in the array:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
%\fill[color=lightgray] (3,0) rectangle (7,1);
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$4$};
|
|
\node at (2.5,0.5) {$4$};
|
|
\node at (3.5,0.5) {$16$};
|
|
\node at (4.5,0.5) {$6$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$4$};
|
|
\node at (7.5,0.5) {$29$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
|
|
\draw[->,thick] (0.5,-0.9) -- (0.5,-0.1);
|
|
\draw[->,thick] (2.5,-0.9) -- (2.5,-0.1);
|
|
\draw[->,thick] (4.5,-0.9) -- (4.5,-0.1);
|
|
\draw[->,thick] (6.5,-0.9) -- (6.5,-0.1);
|
|
\draw[->,thick] (1.5,-1.9) -- (1.5,-0.1);
|
|
\draw[->,thick] (5.5,-1.9) -- (5.5,-0.1);
|
|
\draw[->,thick] (3.5,-2.9) -- (3.5,-0.1);
|
|
\draw[->,thick] (7.5,-3.9) -- (7.5,-0.1);
|
|
|
|
\draw (0,-1) -- (1,-1) -- (1,-1.5) -- (0,-1.5) -- (0,-1);
|
|
\draw (2,-1) -- (3,-1) -- (3,-1.5) -- (2,-1.5) -- (2,-1);
|
|
\draw (4,-1) -- (5,-1) -- (5,-1.5) -- (4,-1.5) -- (4,-1);
|
|
\draw (6,-1) -- (7,-1) -- (7,-1.5) -- (6,-1.5) -- (6,-1);
|
|
\draw (0,-2) -- (2,-2) -- (2,-2.5) -- (0,-2.5) -- (0,-2);
|
|
\draw (4,-2) -- (6,-2) -- (6,-2.5) -- (4,-2.5) -- (4,-2);
|
|
\draw (0,-3) -- (4,-3) -- (4,-3.5) -- (0,-3.5) -- (0,-3);
|
|
\draw (0,-4) -- (8,-4) -- (8,-4.5) -- (0,-4.5) -- (0,-4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
\subsubsection{Sum queries}
|
|
|
|
The values in the binary indexed tree
|
|
can be used to efficiently calculate
|
|
the sum of elements in any range $[1,k]$,
|
|
because such a range
|
|
can be divided into $O(\log n)$ ranges
|
|
whose sums are available in the binary indexed tree.
|
|
|
|
For example, the range $[1,7]$ corresponds to
|
|
the following values:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
%\fill[color=lightgray] (3,0) rectangle (7,1);
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$4$};
|
|
\node at (2.5,0.5) {$4$};
|
|
\node at (3.5,0.5) {$16$};
|
|
\node at (4.5,0.5) {$6$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$4$};
|
|
\node at (7.5,0.5) {$29$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
|
|
\draw[->,thick] (0.5,-0.9) -- (0.5,-0.1);
|
|
\draw[->,thick] (2.5,-0.9) -- (2.5,-0.1);
|
|
\draw[->,thick] (4.5,-0.9) -- (4.5,-0.1);
|
|
\draw[->,thick] (6.5,-0.9) -- (6.5,-0.1);
|
|
\draw[->,thick] (1.5,-1.9) -- (1.5,-0.1);
|
|
\draw[->,thick] (5.5,-1.9) -- (5.5,-0.1);
|
|
\draw[->,thick] (3.5,-2.9) -- (3.5,-0.1);
|
|
\draw[->,thick] (7.5,-3.9) -- (7.5,-0.1);
|
|
|
|
\draw (0,-1) -- (1,-1) -- (1,-1.5) -- (0,-1.5) -- (0,-1);
|
|
\draw (2,-1) -- (3,-1) -- (3,-1.5) -- (2,-1.5) -- (2,-1);
|
|
\draw (4,-1) -- (5,-1) -- (5,-1.5) -- (4,-1.5) -- (4,-1);
|
|
\draw[fill=lightgray] (6,-1) -- (7,-1) -- (7,-1.5) -- (6,-1.5) -- (6,-1);
|
|
\draw (0,-2) -- (2,-2) -- (2,-2.5) -- (0,-2.5) -- (0,-2);
|
|
\draw[fill=lightgray] (4,-2) -- (6,-2) -- (6,-2.5) -- (4,-2.5) -- (4,-2);
|
|
\draw[fill=lightgray] (0,-3) -- (4,-3) -- (4,-3.5) -- (0,-3.5) -- (0,-3);
|
|
\draw (0,-4) -- (8,-4) -- (8,-4.5) -- (0,-4.5) -- (0,-4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Hence, the sum of elements in the range $[1,7]$ is $16+7+4=27$.
|
|
|
|
To calculate the sum of elements in any range $[a,b]$,
|
|
we can use the same trick that we used with prefix sum arrays:
|
|
\[ \textrm{sum}(a,b) = \textrm{sum}(1,b) - \textrm{sum}(1,a-1).\]
|
|
Also in this case, only $O(\log n)$ values are needed.
|
|
|
|
\subsubsection{Array updates}
|
|
|
|
When a value in the array changes,
|
|
several values in the binary indexed tree should be updated.
|
|
For example, if the element at position 3 changes,
|
|
the sums of the following ranges change:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
%\fill[color=lightgray] (3,0) rectangle (7,1);
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$1$};
|
|
\node at (1.5,0.5) {$4$};
|
|
\node at (2.5,0.5) {$4$};
|
|
\node at (3.5,0.5) {$16$};
|
|
\node at (4.5,0.5) {$6$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$4$};
|
|
\node at (7.5,0.5) {$29$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
|
|
\draw[->,thick] (0.5,-0.9) -- (0.5,-0.1);
|
|
\draw[->,thick] (2.5,-0.9) -- (2.5,-0.1);
|
|
\draw[->,thick] (4.5,-0.9) -- (4.5,-0.1);
|
|
\draw[->,thick] (6.5,-0.9) -- (6.5,-0.1);
|
|
\draw[->,thick] (1.5,-1.9) -- (1.5,-0.1);
|
|
\draw[->,thick] (5.5,-1.9) -- (5.5,-0.1);
|
|
\draw[->,thick] (3.5,-2.9) -- (3.5,-0.1);
|
|
\draw[->,thick] (7.5,-3.9) -- (7.5,-0.1);
|
|
|
|
\draw (0,-1) -- (1,-1) -- (1,-1.5) -- (0,-1.5) -- (0,-1);
|
|
\draw[fill=lightgray] (2,-1) -- (3,-1) -- (3,-1.5) -- (2,-1.5) -- (2,-1);
|
|
\draw (4,-1) -- (5,-1) -- (5,-1.5) -- (4,-1.5) -- (4,-1);
|
|
\draw (6,-1) -- (7,-1) -- (7,-1.5) -- (6,-1.5) -- (6,-1);
|
|
\draw (0,-2) -- (2,-2) -- (2,-2.5) -- (0,-2.5) -- (0,-2);
|
|
\draw (4,-2) -- (6,-2) -- (6,-2.5) -- (4,-2.5) -- (4,-2);
|
|
\draw[fill=lightgray] (0,-3) -- (4,-3) -- (4,-3.5) -- (0,-3.5) -- (0,-3);
|
|
\draw[fill=lightgray] (0,-4) -- (8,-4) -- (8,-4.5) -- (0,-4.5) -- (0,-4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Since each array element belongs to $O(\log n)$
|
|
ranges in the binary indexed tree,
|
|
it suffices to update $O(\log n)$ values.
|
|
|
|
|
|
\subsubsection{Implementation}
|
|
|
|
The operations of a binary indexed tree can be implemented
|
|
in an elegant and efficient way using bit operations.
|
|
The key fact needed is that $k \& -k$
|
|
isolates the last one bit of a number $k$.
|
|
For example, $26 \& -26=2$ because the number $26$
|
|
corresponds to 11010 and the number $2$ corresponds to 10.
|
|
|
|
It turns out that when processing a sum query,
|
|
the position $k$ in the binary indexed tree needs to be
|
|
decreased by $k \& -k$ at every step,
|
|
and when updating the array,
|
|
the position $k$ needs to be increased by $k \& -k$ at every step.
|
|
|
|
Suppose that the binary indexed tree is stored in an array \texttt{b}.
|
|
The following function calculates
|
|
the sum of elements in a range $[1,k]$:
|
|
\begin{lstlisting}
|
|
int sum(int k) {
|
|
int s = 0;
|
|
while (k >= 1) {
|
|
s += b[k];
|
|
k -= k&-k;
|
|
}
|
|
return s;
|
|
}
|
|
\end{lstlisting}
|
|
|
|
The following function increases the value
|
|
of the element at position $k$ by $x$
|
|
($x$ can be positive or negative):
|
|
\begin{lstlisting}
|
|
void add(int k, int x) {
|
|
while (k <= n) {
|
|
b[k] += x;
|
|
k += k&-k;
|
|
}
|
|
}
|
|
\end{lstlisting}
|
|
|
|
The time complexity of both the functions is
|
|
$O(\log n)$, because the functions access $O(\log n)$
|
|
values in the binary indexed tree, and each move
|
|
to the next position
|
|
takes $O(1)$ time using bit operations.
|
|
|
|
\section{Segment trees}
|
|
|
|
\index{segment tree}
|
|
|
|
A \key{segment tree}\footnote{The origin of this structure is unknown.
|
|
The bottom-up-implementation in this chapter corresponds to
|
|
the implementation in \cite{sta06}.} is a data structure
|
|
that supports two operations:
|
|
processing a range query and
|
|
modifying an element in the array.
|
|
Segment trees can support
|
|
sum queries, minimum and maximum queries and many other
|
|
queries so that both operations work in $O(\log n)$ time.
|
|
|
|
Compared to a binary indexed tree,
|
|
the advantage of a segment tree is that it is
|
|
a more general data structure.
|
|
While binary indexed trees only support
|
|
sum queries, segment trees also support other queries.
|
|
On the other hand, a segment tree requires more
|
|
memory and is a bit more difficult to implement.
|
|
|
|
\subsubsection{Structure}
|
|
|
|
A segment tree is a binary tree that
|
|
contains $2n-1$ nodes.
|
|
The nodes on the bottom level of the tree
|
|
correspond to the array elements,
|
|
and the other nodes
|
|
contain information needed for processing range queries.
|
|
|
|
Throughout the section, we assume that the size
|
|
of the array is a power of two and zero-based
|
|
indexing is used, because it is convenient to build
|
|
a segment tree for such an array.
|
|
If the size of the array is not a power of two,
|
|
we can always append extra elements to it.
|
|
|
|
We will first discuss segment trees that support sum queries.
|
|
As an example, consider the following array:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$5$};
|
|
\node at (1.5,0.5) {$8$};
|
|
\node at (2.5,0.5) {$6$};
|
|
\node at (3.5,0.5) {$3$};
|
|
\node at (4.5,0.5) {$2$};
|
|
\node at (5.5,0.5) {$7$};
|
|
\node at (6.5,0.5) {$2$};
|
|
\node at (7.5,0.5) {$6$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$0$};
|
|
\node at (1.5,1.4) {$1$};
|
|
\node at (2.5,1.4) {$2$};
|
|
\node at (3.5,1.4) {$3$};
|
|
\node at (4.5,1.4) {$4$};
|
|
\node at (5.5,1.4) {$5$};
|
|
\node at (6.5,1.4) {$6$};
|
|
\node at (7.5,1.4) {$7$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
The corresponding segment tree is as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {5};
|
|
\node[anchor=center] at (1.5, 0.5) {8};
|
|
\node[anchor=center] at (2.5, 0.5) {6};
|
|
\node[anchor=center] at (3.5, 0.5) {3};
|
|
\node[anchor=center] at (4.5, 0.5) {2};
|
|
\node[anchor=center] at (5.5, 0.5) {7};
|
|
\node[anchor=center] at (6.5, 0.5) {2};
|
|
\node[anchor=center] at (7.5, 0.5) {6};
|
|
|
|
\node[draw, circle] (a) at (1,2.5) {13};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,minimum size=22pt] (b) at (3,2.5) {9};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
\node[draw, circle,minimum size=22pt] (c) at (5,2.5) {9};
|
|
\path[draw,thick,-] (c) -- (4.5,1);
|
|
\path[draw,thick,-] (c) -- (5.5,1);
|
|
\node[draw, circle,minimum size=22pt] (d) at (7,2.5) {8};
|
|
\path[draw,thick,-] (d) -- (6.5,1);
|
|
\path[draw,thick,-] (d) -- (7.5,1);
|
|
|
|
\node[draw, circle] (i) at (2,4.5) {22};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\node[draw, circle] (j) at (6,4.5) {17};
|
|
\path[draw,thick,-] (j) -- (c);
|
|
\path[draw,thick,-] (j) -- (d);
|
|
|
|
\node[draw, circle] (m) at (4,6.5) {39};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Each internal node in the segment tree contains
|
|
information about a range of size $2^k$
|
|
in the original array.
|
|
In the above tree, the value of each internal
|
|
node is the sum of the corresponding array elements,
|
|
and it can be calculated as the sum of
|
|
the values of its left and right child node.
|
|
|
|
\subsubsection{Range queries}
|
|
|
|
The sum of elements in a given range
|
|
can be calculated as a sum of values in the segment tree.
|
|
For example, consider the following range:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\fill[color=gray!50] (2,0) rectangle (8,1);
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {5};
|
|
\node[anchor=center] at (1.5, 0.5) {8};
|
|
\node[anchor=center] at (2.5, 0.5) {6};
|
|
\node[anchor=center] at (3.5, 0.5) {3};
|
|
\node[anchor=center] at (4.5, 0.5) {2};
|
|
\node[anchor=center] at (5.5, 0.5) {7};
|
|
\node[anchor=center] at (6.5, 0.5) {2};
|
|
\node[anchor=center] at (7.5, 0.5) {6};
|
|
%
|
|
% \footnotesize
|
|
% \node at (0.5,1.4) {$1$};
|
|
% \node at (1.5,1.4) {$2$};
|
|
% \node at (2.5,1.4) {$3$};
|
|
% \node at (3.5,1.4) {$4$};
|
|
% \node at (4.5,1.4) {$5$};
|
|
% \node at (5.5,1.4) {$6$};
|
|
% \node at (6.5,1.4) {$7$};
|
|
% \node at (7.5,1.4) {$8$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
The sum of elements in the range is
|
|
$6+3+2+7+2+6=26$.
|
|
The following two nodes in the tree
|
|
correspond to the range:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {5};
|
|
\node[anchor=center] at (1.5, 0.5) {8};
|
|
\node[anchor=center] at (2.5, 0.5) {6};
|
|
\node[anchor=center] at (3.5, 0.5) {3};
|
|
\node[anchor=center] at (4.5, 0.5) {2};
|
|
\node[anchor=center] at (5.5, 0.5) {7};
|
|
\node[anchor=center] at (6.5, 0.5) {2};
|
|
\node[anchor=center] at (7.5, 0.5) {6};
|
|
|
|
\node[draw, circle] (a) at (1,2.5) {13};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,fill=gray!50,minimum size=22pt] (b) at (3,2.5) {9};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
\node[draw, circle,minimum size=22pt] (c) at (5,2.5) {9};
|
|
\path[draw,thick,-] (c) -- (4.5,1);
|
|
\path[draw,thick,-] (c) -- (5.5,1);
|
|
\node[draw, circle,minimum size=22pt] (d) at (7,2.5) {8};
|
|
\path[draw,thick,-] (d) -- (6.5,1);
|
|
\path[draw,thick,-] (d) -- (7.5,1);
|
|
|
|
\node[draw, circle] (i) at (2,4.5) {22};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\node[draw, circle,fill=gray!50] (j) at (6,4.5) {17};
|
|
\path[draw,thick,-] (j) -- (c);
|
|
\path[draw,thick,-] (j) -- (d);
|
|
|
|
\node[draw, circle] (m) at (4,6.5) {39};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
Thus, the sum of elements in the range is $9+17=26$.
|
|
|
|
When the sum is calculated using nodes
|
|
that are located as high as possible in the tree,
|
|
at most two nodes on each level
|
|
of the tree are needed.
|
|
Hence, the total number of nodes
|
|
is only $O(\log n)$.
|
|
|
|
\subsubsection{Array updates}
|
|
|
|
When an element in the array changes,
|
|
we should update all nodes in the tree
|
|
whose value depends on the element.
|
|
This can be done by traversing the path
|
|
from the element to the top node
|
|
and updating the nodes along the path.
|
|
|
|
\begin{samepage}
|
|
The following picture shows which nodes in the segment tree
|
|
change if the element 7 in the array changes.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\fill[color=gray!50] (5,0) rectangle (6,1);
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {5};
|
|
\node[anchor=center] at (1.5, 0.5) {8};
|
|
\node[anchor=center] at (2.5, 0.5) {6};
|
|
\node[anchor=center] at (3.5, 0.5) {3};
|
|
\node[anchor=center] at (4.5, 0.5) {2};
|
|
\node[anchor=center] at (5.5, 0.5) {7};
|
|
\node[anchor=center] at (6.5, 0.5) {2};
|
|
\node[anchor=center] at (7.5, 0.5) {6};
|
|
|
|
\node[draw, circle] (a) at (1,2.5) {13};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,minimum size=22pt] (b) at (3,2.5) {9};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
\node[draw, circle,minimum size=22pt,fill=gray!50] (c) at (5,2.5) {9};
|
|
\path[draw,thick,-] (c) -- (4.5,1);
|
|
\path[draw,thick,-] (c) -- (5.5,1);
|
|
\node[draw, circle,minimum size=22pt] (d) at (7,2.5) {8};
|
|
\path[draw,thick,-] (d) -- (6.5,1);
|
|
\path[draw,thick,-] (d) -- (7.5,1);
|
|
|
|
\node[draw, circle] (i) at (2,4.5) {22};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\node[draw, circle,fill=gray!50] (j) at (6,4.5) {17};
|
|
\path[draw,thick,-] (j) -- (c);
|
|
\path[draw,thick,-] (j) -- (d);
|
|
|
|
\node[draw, circle,fill=gray!50] (m) at (4,6.5) {39};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
The path from bottom to top
|
|
always consists of $O(\log n)$ nodes,
|
|
so each update changes $O(\log n)$ nodes in the tree.
|
|
|
|
\subsubsection{Storing the tree}
|
|
|
|
A segment tree can be stored in an array
|
|
of $2N$ elements where $N$ is a power of two.
|
|
Such a tree corresponds to an array
|
|
indexed from $0$ to $N-1$.
|
|
|
|
In the segment tree array,
|
|
the element at position 1
|
|
corresponds to the top node of the tree,
|
|
the elements at positions 2 and 3 correspond to
|
|
the second level of the tree, and so on.
|
|
Finally, the elements at positions $N \ldots 2N-1$
|
|
correspond to the bottom level of the tree, i.e.,
|
|
the elements of the original array.
|
|
|
|
For example, the segment tree
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {5};
|
|
\node[anchor=center] at (1.5, 0.5) {8};
|
|
\node[anchor=center] at (2.5, 0.5) {6};
|
|
\node[anchor=center] at (3.5, 0.5) {3};
|
|
\node[anchor=center] at (4.5, 0.5) {2};
|
|
\node[anchor=center] at (5.5, 0.5) {7};
|
|
\node[anchor=center] at (6.5, 0.5) {2};
|
|
\node[anchor=center] at (7.5, 0.5) {6};
|
|
|
|
\node[draw, circle] (a) at (1,2.5) {13};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,minimum size=22pt] (b) at (3,2.5) {9};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
\node[draw, circle,minimum size=22pt] (c) at (5,2.5) {9};
|
|
\path[draw,thick,-] (c) -- (4.5,1);
|
|
\path[draw,thick,-] (c) -- (5.5,1);
|
|
\node[draw, circle,minimum size=22pt] (d) at (7,2.5) {8};
|
|
\path[draw,thick,-] (d) -- (6.5,1);
|
|
\path[draw,thick,-] (d) -- (7.5,1);
|
|
|
|
\node[draw, circle] (i) at (2,4.5) {22};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\node[draw, circle] (j) at (6,4.5) {17};
|
|
\path[draw,thick,-] (j) -- (c);
|
|
\path[draw,thick,-] (j) -- (d);
|
|
|
|
\node[draw, circle] (m) at (4,6.5) {39};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
can be stored as follows ($N=8$):
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
%\fill[color=lightgray] (3,0) rectangle (7,1);
|
|
\draw (0,0) grid (15,1);
|
|
|
|
\node at (0.5,0.5) {$39$};
|
|
\node at (1.5,0.5) {$22$};
|
|
\node at (2.5,0.5) {$17$};
|
|
\node at (3.5,0.5) {$13$};
|
|
\node at (4.5,0.5) {$9$};
|
|
\node at (5.5,0.5) {$9$};
|
|
\node at (6.5,0.5) {$8$};
|
|
\node at (7.5,0.5) {$5$};
|
|
\node at (8.5,0.5) {$8$};
|
|
\node at (9.5,0.5) {$6$};
|
|
\node at (10.5,0.5) {$3$};
|
|
\node at (11.5,0.5) {$2$};
|
|
\node at (12.5,0.5) {$7$};
|
|
\node at (13.5,0.5) {$2$};
|
|
\node at (14.5,0.5) {$6$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\node at (8.5,1.4) {$9$};
|
|
\node at (9.5,1.4) {$10$};
|
|
\node at (10.5,1.4) {$11$};
|
|
\node at (11.5,1.4) {$12$};
|
|
\node at (12.5,1.4) {$13$};
|
|
\node at (13.5,1.4) {$14$};
|
|
\node at (14.5,1.4) {$15$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
Using this representation,
|
|
for a node at position $k$,
|
|
\begin{itemize}
|
|
\item the parent node is at position $\lfloor k/2 \rfloor$,
|
|
\item the left child node is at position $2k$, and
|
|
\item the right child node is at position $2k+1$.
|
|
\end{itemize}
|
|
% Note that this implies that the index of a node
|
|
% is even if it is a left child and odd if it is a right child.
|
|
|
|
\subsubsection{Functions}
|
|
|
|
Assume that the segment tree is stored
|
|
in an array \texttt{p}.
|
|
The following function
|
|
calculates the sum of elements in a range $[a,b]$:
|
|
|
|
\begin{lstlisting}
|
|
int sum(int a, int b) {
|
|
a += N; b += N;
|
|
int s = 0;
|
|
while (a <= b) {
|
|
if (a%2 == 1) s += p[a++];
|
|
if (b%2 == 0) s += p[b--];
|
|
a /= 2; b /= 2;
|
|
}
|
|
return s;
|
|
}
|
|
\end{lstlisting}
|
|
|
|
The function starts at the bottom of the tree
|
|
and moves one level up at each step.
|
|
Initially, the range $[a+N,b+N]$ corresponds
|
|
to the range $[a,b]$ in the original array.
|
|
At each step, the function adds the value of
|
|
the left and right node to the sum
|
|
if their parent nodes do not belong to the range.
|
|
This process continues, until the sum of the
|
|
range has been calculated.
|
|
|
|
The following function increases the value
|
|
of the element at position $k$ by $x$:
|
|
|
|
\begin{lstlisting}
|
|
void add(int k, int x) {
|
|
k += N;
|
|
p[k] += x;
|
|
for (k /= 2; k >= 1; k /= 2) {
|
|
p[k] = p[2*k]+p[2*k+1];
|
|
}
|
|
}
|
|
\end{lstlisting}
|
|
First the function updates the element
|
|
at the bottom level of the tree.
|
|
After this, the function updates the values of all
|
|
internal nodes in the tree, until it reaches
|
|
the top node of the tree.
|
|
|
|
Both above functions work
|
|
in $O(\log n)$ time, because a segment tree
|
|
of $n$ elements consists of $O(\log n)$ levels,
|
|
and the operations move one level forward in the tree at each step.
|
|
|
|
\subsubsection{Other queries}
|
|
|
|
Segment trees can support any queries
|
|
as long as we can divide a range into two parts,
|
|
calculate the answer for both parts
|
|
and then efficiently combine the answers.
|
|
Examples of such queries are
|
|
minimum and maximum, greatest common divisor,
|
|
and bit operations and, or and xor.
|
|
|
|
\begin{samepage}
|
|
For example, the following segment tree
|
|
supports minimum queries:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {5};
|
|
\node[anchor=center] at (1.5, 0.5) {8};
|
|
\node[anchor=center] at (2.5, 0.5) {6};
|
|
\node[anchor=center] at (3.5, 0.5) {3};
|
|
\node[anchor=center] at (4.5, 0.5) {1};
|
|
\node[anchor=center] at (5.5, 0.5) {7};
|
|
\node[anchor=center] at (6.5, 0.5) {2};
|
|
\node[anchor=center] at (7.5, 0.5) {6};
|
|
|
|
\node[draw, circle,minimum size=22pt] (a) at (1,2.5) {5};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,minimum size=22pt] (b) at (3,2.5) {3};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
\node[draw, circle,minimum size=22pt] (c) at (5,2.5) {1};
|
|
\path[draw,thick,-] (c) -- (4.5,1);
|
|
\path[draw,thick,-] (c) -- (5.5,1);
|
|
\node[draw, circle,minimum size=22pt] (d) at (7,2.5) {2};
|
|
\path[draw,thick,-] (d) -- (6.5,1);
|
|
\path[draw,thick,-] (d) -- (7.5,1);
|
|
|
|
\node[draw, circle,minimum size=22pt] (i) at (2,4.5) {3};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\node[draw, circle,minimum size=22pt] (j) at (6,4.5) {1};
|
|
\path[draw,thick,-] (j) -- (c);
|
|
\path[draw,thick,-] (j) -- (d);
|
|
|
|
\node[draw, circle,minimum size=22pt] (m) at (4,6.5) {1};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
In this segment tree, every node in the tree
|
|
contains the smallest element in the corresponding
|
|
range of the array.
|
|
The top node of the tree contains the smallest
|
|
element in the whole array.
|
|
The operations can be implemented like previously,
|
|
but instead of sums, minima are calculated.
|
|
|
|
\subsubsection{Binary search in a tree}
|
|
|
|
The structure of the segment tree allows us
|
|
to use binary search for finding elements in the array.
|
|
For example, if the tree supports minimum queries,
|
|
we can find the position of the smallest
|
|
element in $O(\log n)$ time.
|
|
|
|
For example, in the following tree the
|
|
smallest element 1 can be found
|
|
by traversing a path downwards from the top node:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (8,0) grid (16,1);
|
|
|
|
\node[anchor=center] at (8.5, 0.5) {9};
|
|
\node[anchor=center] at (9.5, 0.5) {5};
|
|
\node[anchor=center] at (10.5, 0.5) {7};
|
|
\node[anchor=center] at (11.5, 0.5) {1};
|
|
\node[anchor=center] at (12.5, 0.5) {6};
|
|
\node[anchor=center] at (13.5, 0.5) {2};
|
|
\node[anchor=center] at (14.5, 0.5) {3};
|
|
\node[anchor=center] at (15.5, 0.5) {2};
|
|
|
|
%\node[anchor=center] at (1,2.5) {13};
|
|
|
|
\node[draw, circle,minimum size=22pt] (e) at (9,2.5) {5};
|
|
\path[draw,thick,-] (e) -- (8.5,1);
|
|
\path[draw,thick,-] (e) -- (9.5,1);
|
|
\node[draw, circle,minimum size=22pt] (f) at (11,2.5) {1};
|
|
\path[draw,thick,-] (f) -- (10.5,1);
|
|
\path[draw,thick,-] (f) -- (11.5,1);
|
|
\node[draw, circle,minimum size=22pt] (g) at (13,2.5) {2};
|
|
\path[draw,thick,-] (g) -- (12.5,1);
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\path[draw,thick,-] (g) -- (13.5,1);
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\node[draw, circle,minimum size=22pt] (h) at (15,2.5) {2};
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\path[draw,thick,-] (h) -- (14.5,1);
|
|
\path[draw,thick,-] (h) -- (15.5,1);
|
|
|
|
\node[draw, circle,minimum size=22pt] (k) at (10,4.5) {1};
|
|
\path[draw,thick,-] (k) -- (e);
|
|
\path[draw,thick,-] (k) -- (f);
|
|
\node[draw, circle,minimum size=22pt] (l) at (14,4.5) {2};
|
|
\path[draw,thick,-] (l) -- (g);
|
|
\path[draw,thick,-] (l) -- (h);
|
|
|
|
\node[draw, circle,minimum size=22pt] (n) at (12,6.5) {1};
|
|
\path[draw,thick,-] (n) -- (k);
|
|
\path[draw,thick,-] (n) -- (l);
|
|
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (n) -- (k);
|
|
\path[draw=red,thick,->,line width=2pt] (k) -- (f);
|
|
\path[draw=red,thick,->,line width=2pt] (f) -- (11.5,1);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
\section{Additional techniques}
|
|
|
|
\subsubsection{Index compression}
|
|
|
|
A limitation in data structures that
|
|
are built upon an array is that
|
|
the elements are indexed using
|
|
consecutive integers.
|
|
Difficulties arise when large indices
|
|
are needed.
|
|
For example, if we wish to use the index $10^9$,
|
|
the array should contain $10^9$
|
|
elements which would require too much memory.
|
|
|
|
\index{index compression}
|
|
|
|
However, we can often bypass this limitation
|
|
by using \key{index compression},
|
|
where the original indices are replaced
|
|
with indices $1,2,3,$ etc.
|
|
This can be done if we know all the indices
|
|
needed during the algorithm beforehand.
|
|
|
|
The idea is to replace each original index $x$
|
|
with $p(x)$ where $p$ is a function that
|
|
compresses the indices.
|
|
We require that the order of the indices
|
|
does not change, so if $a<b$, then $p(a)<p(b)$.
|
|
This allows us to conviently perform queries
|
|
even if the indices are compressed.
|
|
|
|
For example, if the original indices are
|
|
$555$, $10^9$ and $8$, the new indices are:
|
|
|
|
\[
|
|
\begin{array}{lcl}
|
|
p(8) & = & 1 \\
|
|
p(555) & = & 2 \\
|
|
p(10^9) & = & 3 \\
|
|
\end{array}
|
|
\]
|
|
|
|
\subsubsection{Range updates}
|
|
|
|
So far, we have implemented data structures
|
|
that support range queries and updates
|
|
of single values.
|
|
Let us now consider an opposite situation,
|
|
where we should update ranges and
|
|
retrieve single values.
|
|
We focus on an operation that increases all
|
|
elements in a range $[a,b]$ by $x$.
|
|
|
|
\index{difference array}
|
|
|
|
Surprisingly, we can use the data structures
|
|
presented in this chapter also in this situation.
|
|
To do this, we build a \key{difference array}
|
|
for the array.
|
|
In such an array, each value indicates
|
|
the difference between two consecutive values
|
|
in the original array.
|
|
Thus, the original array is the
|
|
prefix sum array of the
|
|
difference array.
|
|
For example, consider the following array:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$3$};
|
|
\node at (1.5,0.5) {$3$};
|
|
\node at (2.5,0.5) {$1$};
|
|
\node at (3.5,0.5) {$1$};
|
|
\node at (4.5,0.5) {$1$};
|
|
\node at (5.5,0.5) {$5$};
|
|
\node at (6.5,0.5) {$2$};
|
|
\node at (7.5,0.5) {$2$};
|
|
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The difference array for the above array is as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$3$};
|
|
\node at (1.5,0.5) {$0$};
|
|
\node at (2.5,0.5) {$-2$};
|
|
\node at (3.5,0.5) {$0$};
|
|
\node at (4.5,0.5) {$0$};
|
|
\node at (5.5,0.5) {$4$};
|
|
\node at (6.5,0.5) {$-3$};
|
|
\node at (7.5,0.5) {$0$};
|
|
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
For example, the value 5 at position 6 in the original array
|
|
corresponds to the sum $3-2+4=5$.
|
|
|
|
The advantage of the difference array is
|
|
that we can update a range
|
|
in the original array by changing just
|
|
two elements in the difference array.
|
|
For example, if we want to
|
|
increase the elements in the range $2 \ldots 5$ by 5,
|
|
it suffices to increase the value at position 2 by 5
|
|
and decrease the value at position 6 by 5.
|
|
The result is as follows:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node at (0.5,0.5) {$3$};
|
|
\node at (1.5,0.5) {$5$};
|
|
\node at (2.5,0.5) {$-2$};
|
|
\node at (3.5,0.5) {$0$};
|
|
\node at (4.5,0.5) {$0$};
|
|
\node at (5.5,0.5) {$-1$};
|
|
\node at (6.5,0.5) {$-3$};
|
|
\node at (7.5,0.5) {$0$};
|
|
|
|
\footnotesize
|
|
\node at (0.5,1.4) {$1$};
|
|
\node at (1.5,1.4) {$2$};
|
|
\node at (2.5,1.4) {$3$};
|
|
\node at (3.5,1.4) {$4$};
|
|
\node at (4.5,1.4) {$5$};
|
|
\node at (5.5,1.4) {$6$};
|
|
\node at (6.5,1.4) {$7$};
|
|
\node at (7.5,1.4) {$8$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
More generally, to increase the elements
|
|
in a range $[a,b]$ by $x$,
|
|
we increase the value at position $a$ by $x$
|
|
and decrease the value at position $b+1$ by $x$.
|
|
Thus, it is only needed to update single values
|
|
and process sum queries,
|
|
so we can use a binary indexed tree or a segment tree.
|
|
|
|
A more difficult problem is to support both
|
|
range queries and range updates.
|
|
In Chapter 28 we will see that even this is possible.
|
|
|
|
|
|
|