1424 lines
40 KiB
TeX
1424 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 a \key{range query}, a range of an array
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is given and we should calculate some value from the
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elements in the range. Typical range queries are:
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\begin{itemize}
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\item \key{sum query}: calculate the sum of elements in range $[a,b]$
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\item \key{minimum query}: find the smallest element in range $[a,b]$
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\item \key{maximum query}: find the largest element in range $[a,b]$
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\end{itemize}
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For example, in range $[4,7]$ of the following array,
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the sum is $4+6+1+3=14$, the minimum is 1 and the maximum is 6:
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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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An easy way to answer a range query is
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to iterate through all the elements in the range.
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For example, we can answer a sum query as follows:
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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 handles a sum query
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in $O(n)$ time, which is slow if the array is large
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and there are a lot of queries.
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In this chapter we will learn how
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range queries can be answered much more efficiently.
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\section{Static array queries}
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We will first focus on a simple case where
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the array is \key{static}, i.e.,
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the elements never change between the queries.
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In this case, it suffices to process the
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contents of the array beforehand and construct
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a data structure that can be used for answering
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any possible range query efficiently.
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\subsubsection{Sum query}
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\index{prefix sum array}
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Sum queries can be answered efficiently
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by constructing a \key{sum array}
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that contains the sum of the range $[1,k]$
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for each $k=1,2,\ldots,n$.
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After this, the sum of any range $[a,b]$ of the
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original array
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can be calculated in $O(1)$ time using the
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precalculated sum array.
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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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%\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 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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The following code constructs a prefix sum
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array \texttt{s} from array \texttt{t} in $O(n)$ time:
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\begin{lstlisting}
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for (int i = 1; i <= n; i++) {
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s[i] = s[i-1]+t[i];
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}
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\end{lstlisting}
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After this, the following function answers
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a sum query in $O(1)$ time:
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\begin{lstlisting}
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int sum(int a, int b) {
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return s[b]-s[a-1];
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}
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\end{lstlisting}
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The function calculates the sum of range $[a,b]$
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by subtracting the sum of range $[1,a-1]$
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from the sum of range $[1,b]$.
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Thus, only two values from the sum array
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are needed, and the query takes $O(1)$ time.
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Note that thanks to the one-based indexing,
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the function also works when $a=1$ if $\texttt{s}[0]=0$.
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As an 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 of the range $[4,7]$ is $8+6+1+4=19$.
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This can be calculated from the sum array
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using the sums $[1,3]$ and $[1,7]$:
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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 of the range $[4,7]$ is $27-8=19$.
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We can also generalize the idea of a sum array
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for a two-dimensional array.
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In this case, it will be possible to calculate the sum of
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any rectangular subarray in $O(1)$ time.
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The sum array will contain sums
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for all subarrays that begin from the upper-left corner.
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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=0.55]
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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 inside 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 in a rectangular
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subarray from the upper-left corner
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to the position of letter $X$.
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\subsubsection{Minimum query}
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It is also possible to answer a minimum query
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in $O(1)$ time after preprocessing, though it is
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more difficult than answer a sum query.
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Note that minimum and maximum queries can always
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be implemented using same techniques,
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so it suffices to focus on the minimum query.
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The idea is to find the minimum element for each range
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of size $2^k$ in the array.
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For example, in 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 minima 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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range & size & min \\
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\hline
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$[1,1]$ & 1 & 1 \\
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$[2,2]$ & 1 & 3 \\
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$[3,3]$ & 1 & 4 \\
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$[4,4]$ & 1 & 8 \\
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$[5,5]$ & 1 & 6 \\
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$[6,6]$ & 1 & 1 \\
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$[7,7]$ & 1 & 4 \\
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$[8,8]$ & 1 & 2 \\
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\end{tabular}
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&
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\begin{tabular}{ccc}
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range & size & min \\
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\hline
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$[1,2]$ & 2 & 1 \\
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$[2,3]$ & 2 & 3 \\
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$[3,4]$ & 2 & 4 \\
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$[4,5]$ & 2 & 6 \\
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$[5,6]$ & 2 & 1 \\
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$[6,7]$ & 2 & 1 \\
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$[7,8]$ & 2 & 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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range & size & min \\
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\hline
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$[1,4]$ & 4 & 1 \\
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$[2,5]$ & 4 & 3 \\
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$[3,6]$ & 4 & 1 \\
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$[4,7]$ & 4 & 1 \\
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$[5,8]$ & 4 & 1 \\
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$[1,8]$ & 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 $2^k$ ranges in an array is $O(n \log n)$
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because there are $O(\log n)$ ranges that begin
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from each array index.
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The minima for all $2^k$ ranges can be calculated
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in $O(n \log n)$ time because each $2^k$ range
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consists of two $2^{k-1}$ ranges, so the minima
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can be calculated recursively.
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After this, the minimum of any range $[a,b]$c
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can be calculated in $O(1)$ time as a minimum of
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two $2^k$ ranges where $k=\lfloor \log_2(b-a+1) \rfloor$.
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The first range begins from index $a$,
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and the second range ends to index $b$.
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The parameter $k$ is so chosen that
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two $2^k$ ranges cover the range $[a,b]$ entirely.
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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 $[2,7]$ is 6,
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and $\lfloor \log_2(6) \rfloor = 2$.
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Thus, the minimum can be calculated
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from two ranges of length 4.
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The ranges are $[2,5]$ 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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\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 minimum of the range $[2,5]$ is 3,
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and the minimum of the range $[4,7]$ is 1.
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Thus, the minimum of the range $[2,7]$ is 1.
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\section{Binary indexed tree}
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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 a \key{Fenwick tree}
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|
is a data structure that resembles a sum array.
|
|
The supported operations are answering
|
|
a sum query for range $[a,b]$,
|
|
and updating the element at index $k$.
|
|
The time complexity for both of the operations is $O(\log n)$.
|
|
|
|
Unlike a sum array, a binary indexed tree
|
|
can be efficiently updated between the sum queries.
|
|
This would not be possible using a sum array
|
|
because we should build the whole sum array again
|
|
in $O(n)$ time after each update.
|
|
|
|
\subsubsection{Structure}
|
|
|
|
A binary indexed tree can be represented as an array
|
|
where index $k$ contains the sum of a range in the
|
|
original array that ends to index $k$.
|
|
The length of the range is the largest power of two
|
|
that divides $k$.
|
|
For example, if $k=6$, the length of the range is $2$
|
|
because $2$ divides $6$ but $4$ doesn't divide $6$.
|
|
|
|
\begin{samepage}
|
|
For example, for the 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}
|
|
the corresponding binary indexed tree is as follows:
|
|
\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}
|
|
|
|
For example, the binary indexed tree
|
|
contains the value 7 at index 6
|
|
because the sum of the elements in the range $[5,6]$
|
|
of the original array is $6+1=7$.
|
|
|
|
\subsubsection{Sum query}
|
|
|
|
The basic operation in a binary indexed tree is
|
|
calculating the sum of a range $[1,k]$ where $k$
|
|
is any index in the array.
|
|
The sum of any range can be constructed by combining
|
|
sums of subranges in the tree.
|
|
|
|
For example, the range $[1,7]$ will be divided
|
|
into three subranges:
|
|
\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}
|
|
|
|
Thus, the sum of the range $[1,7]$ is $16+7+4=27$.
|
|
Because of the structure of the binary indexed tree,
|
|
the length of each subrange inside a range is distinct,
|
|
so the sum of a range
|
|
always consists of sums of $O(\log n)$ subranges.
|
|
|
|
Using the same technique that we previously used
|
|
with a sum array,
|
|
we can efficiently calculate the sum of any range
|
|
$[a,b]$ by substracting the sum of the range $[1,a-1]$
|
|
from the sum of the range $[1,b]$.
|
|
The time complexity remains $O(\log n)$
|
|
because it suffices to calculate two sums of $[1,k]$ ranges.
|
|
|
|
\subsubsection{Array update}
|
|
|
|
When an element in the original array changes,
|
|
several sums in the binary indexed tree change.
|
|
For example, if the value at index 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}
|
|
|
|
Also in this case, the length of each range is distinct,
|
|
so $O(\log n)$ ranges will be updated in the binary indexed tree.
|
|
|
|
\subsubsection{Implementation}
|
|
|
|
The operations of a binary indexed tree can be implemented
|
|
in an elegant and efficient way using bit manipulation.
|
|
The bit operation needed is $k \& -k$ that
|
|
returns the last bit one from number $k$.
|
|
For example, $6 \& -6=2$ because the number $6$
|
|
corresponds to 110 and the number $2$ corresponds to 10.
|
|
|
|
It turns out that when calculating a range sum,
|
|
the index $k$ in the binary indexed tree should be
|
|
decreased by $k \& -k$ at every step.
|
|
Correspondingly, when updating the array,
|
|
the index $k$ should be increased by $k \& -k$ at every step.
|
|
|
|
The following functions assume that the binary indexed tree
|
|
is stored to array \texttt{b} and it consists of indices $1 \ldots n$.
|
|
|
|
The function \texttt{sum} calculates the sum of the 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 function \texttt{add} increases the value of element $k$ by $x$:
|
|
\begin{lstlisting}
|
|
void add(int k, int x) {
|
|
while (k <= n) {
|
|
b[k] += x;
|
|
k += k&-k;
|
|
}
|
|
}
|
|
\end{lstlisting}
|
|
|
|
The time complexity of both above functions is
|
|
$O(\log n)$ because the functions change $O(\log n)$
|
|
values in the binary indexed tree and each move
|
|
to the next index
|
|
takes $O(1)$ time using the bit operation.
|
|
|
|
\section{Segmenttipuu}
|
|
|
|
\index{segmenttipuu@segmenttipuu}
|
|
|
|
\key{Segmenttipuu} on tietorakenne,
|
|
jonka operaatiot ovat taulukon välin $[a,b]$ välikysely
|
|
sekä kohdan $k$ arvon päivitys.
|
|
Segmenttipuun avulla voi toteuttaa summakyselyn,
|
|
minimikyselyn ja monia muitakin kyselyitä niin,
|
|
että kummankin operaation aikavaativuus on $O(\log n)$.
|
|
|
|
Segmenttipuun etuna binääri-indeksipuuhun verrattuna on,
|
|
että se on yleisempi tietorakenne.
|
|
Binääri-indeksipuulla voi toteuttaa vain summakyselyn,
|
|
mutta segmenttipuu sallii muitakin kyselyitä.
|
|
Toisaalta segmenttipuu vie enemmän muistia ja
|
|
on hieman vaikeampi toteuttaa kuin binääri-indeksipuu.
|
|
|
|
\subsubsection{Rakenne}
|
|
|
|
Segmenttipuussa on $2n-1$ solmua niin,
|
|
että alimmalla tasolla on $n$ solmua,
|
|
jotka kuvaavat taulukon sisällön,
|
|
ja ylemmillä tasoilla on välikyselyihin
|
|
tarvittavaa tietoa.
|
|
Segmenttipuun sisältö riippuu siitä,
|
|
mikä välikysely puun tulee toteuttaa.
|
|
Oletamme aluksi, että välikysely on tuttu summakysely.
|
|
|
|
Esimerkiksi taulukkoa
|
|
\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) {$1$};
|
|
\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}
|
|
vastaa seuraava segmenttipuu:
|
|
\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}
|
|
|
|
Jokaisessa segmenttipuun solmussa on tietoa
|
|
$2^k$-kokoisesta välistä taulukossa.
|
|
Tässä tapauksessa solmussa oleva arvo kertoo,
|
|
mikä on taulukon lukujen summa solmua vastaavalla
|
|
välillä.
|
|
Kunkin solmun arvo saadaan laskemalla yhteen
|
|
solmun alapuolella vasemmalla ja oikealla
|
|
olevien solmujen arvot.
|
|
|
|
Segmenttipuu on mukavinta rakentaa niin,
|
|
että taulukon koko on 2:n potenssi,
|
|
jolloin tuloksena on täydellinen binääripuu.
|
|
Jatkossa oletamme aina,
|
|
että taulukko täyttää tämän vaatimuksen.
|
|
Jos taulukon koko ei ole 2:n potenssi,
|
|
sen loppuun voi lisätä tyhjää niin,
|
|
että koosta tulee 2:n potenssi.
|
|
|
|
\subsubsection{Välikysely}
|
|
|
|
Segmenttipuussa vastaus välikyselyyn lasketaan
|
|
väliin kuuluvista solmuista,
|
|
jotka ovat mahdollisimman korkealla puussa.
|
|
Jokainen solmu antaa vastauksen väliin kuuluvalle osavälille,
|
|
ja vastaus kyselyyn selviää yhdistämällä
|
|
segmenttipuusta saadut osavälejä koskeva tiedot.
|
|
|
|
Tarkastellaan esimerkiksi seuraavaa taulukon väliä:
|
|
\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}
|
|
Lukujen summa välillä $[3,8]$ on $6+3+2+7+2+6=26$.
|
|
Segmenttipuusta summa saadaan laskettua seuraavien
|
|
osasummien avulla:
|
|
\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}
|
|
Taulukon välin summaksi tulee osasummista $9+17=26$.
|
|
|
|
Kun vastaus välikyselyyn lasketaan mahdollisimman
|
|
korkealla segmenttipuussa olevista solmuista,
|
|
väliin kuuluu enintään kaksi solmua
|
|
jokaiselta segmenttipuun tasolta.
|
|
Tämän ansiosta välikyselyssä
|
|
tarvittavien solmujen yhteismäärä on vain $O(\log n)$.
|
|
|
|
\subsubsection{Taulukon päivitys}
|
|
|
|
Kun taulukossa oleva arvo muuttuu,
|
|
segmenttipuussa täytyy päivittää
|
|
kaikkia solmuja, joiden arvo
|
|
riippuu muutetusta taulukon kohdasta.
|
|
Tämä tapahtuu kulkemalla puuta ylöspäin huipulle
|
|
asti ja tekemällä muutokset.
|
|
|
|
\begin{samepage}
|
|
Seuraava kuva näyttää, mitkä solmut segmenttipuussa muuttuvat,
|
|
jos taulukon luku 7 muuttuu.
|
|
|
|
\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}
|
|
|
|
Polku segmenttipuun pohjalta huipulle muodostuu aina $O(\log n)$ solmusta,
|
|
joten taulukon arvon muuttuminen vaikuttaa $O(\log n)$ solmuun puussa.
|
|
|
|
\subsubsection{Puun tallennus}
|
|
|
|
Segmenttipuun voi tallentaa muistiin
|
|
$2N$ alkion taulukkona,
|
|
jossa $N$ on riittävän suuri 2:n potenssi.
|
|
Tällaisen segmenttipuun avulla voi ylläpitää
|
|
taulukkoa, jonka indeksialue on $[0,N-1]$.
|
|
|
|
Segmenttipuun taulukon
|
|
kohdassa 1 on puun ylimmän solmun arvo,
|
|
kohdat 2 ja 3 sisältävät seuraavan tason
|
|
solmujen arvot, jne.
|
|
Segmenttipuun alin taso eli varsinainen
|
|
taulukon sisältä tallennetaan
|
|
kohdasta $N$ alkaen.
|
|
Niinpä taulukon kohdassa $k$ oleva alkio
|
|
on segmenttipuun taulukossa kohdassa $k+N$.
|
|
|
|
Esimerkiksi segmenttipuun
|
|
\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}
|
|
voi tallentaa taulukkoon seuraavasti ($N=8$):
|
|
\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}
|
|
Tätä tallennustapaa käyttäen kohdassa $k$
|
|
olevalle solmulle pätee, että
|
|
\begin{itemize}
|
|
\item ylempi solmu on kohdassa $\lfloor k/2 \rfloor$,
|
|
\item vasen alempi solmu on kohdassa $2k$ ja
|
|
\item oikea alempi solmu on kohdassa $2k+1$.
|
|
\end{itemize}
|
|
Huomaa, että tämän seurauksena solmun kohta on parillinen,
|
|
jos se on vasemmalla ylemmästä solmusta katsoen,
|
|
ja pariton, jos se on oikealla.
|
|
|
|
\subsubsection{Toteutus}
|
|
|
|
Tarkastellaan seuraavaksi välikyselyn ja päivityksen
|
|
toteutusta segmenttipuuhun.
|
|
Seuraavat funktiot olettavat, että segmenttipuu
|
|
on tallennettu $2n-1$-kokoi\-seen taulukkoon $\texttt{p}$
|
|
edellä kuvatulla tavalla.
|
|
|
|
Funktio \texttt{summa} laskee summan
|
|
välillä $a \ldots b$:
|
|
|
|
\begin{lstlisting}
|
|
int summa(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}
|
|
|
|
Funktio aloittaa summan laskeminen segmenttipuun
|
|
pohjalta ja liikkuu askel kerrallaan ylemmille tasoille.
|
|
Funktio laskee välin summan muuttujaan $s$
|
|
yhdistämällä puussa olevia osasummia.
|
|
Välin reunalla oleva osasumma lisätään summaan
|
|
aina silloin, kun se ei kuulu ylemmän tason osasummaan.
|
|
|
|
Funktio \texttt{lisaa} kasvattaa kohdan $k$ arvoa $x$:llä:
|
|
|
|
\begin{lstlisting}
|
|
void lisaa(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}
|
|
Ensin funktio tekee muutoksen puun alimmalle
|
|
tasolle taulukkoon.
|
|
Tämän jälkeen se päivittää kaikki osasummat
|
|
puun huipulle asti.
|
|
Taulukon \texttt{p} indeksoinnin ansiosta
|
|
kohdasta $k$ alemmalla tasolla
|
|
ovat kohdat $2k$ ja $2k+1$.
|
|
|
|
Molemmat segmenttipuun operaatiot toimivat ajassa
|
|
$O(\log n)$, koska $n$ lukua sisältävässä
|
|
segmenttipuussa on $O(\log n)$ tasoa
|
|
ja operaatiot siirtyvät askel kerrallaan
|
|
segmenttipuun tasoja ylöspäin.
|
|
|
|
\subsubsection{Muut kyselyt}
|
|
|
|
Segmenttipuu mahdollistaa summan lisäksi minkä
|
|
tahansa välikyselyn,
|
|
jossa vierekkäisten välien $[a,b]$ ja $[b+1,c]$
|
|
tuloksista pystyy laskemaan tehokkaasti
|
|
välin $[a,c]$ tuloksen.
|
|
Tällaisia kyselyitä
|
|
ovat esimerkiksi minimi ja maksimi,
|
|
suurin yhteinen tekijä
|
|
sekä bittioperaatiot and, or ja xor.
|
|
|
|
\begin{samepage}
|
|
Esimerkiksi seuraavan segmenttipuun avulla voi laskea
|
|
taulukon välien minimejä:
|
|
|
|
\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}
|
|
|
|
Tässä segmenttipuussa jokainen puun solmu kertoo,
|
|
mikä on pienin luku sen alapuolella olevassa
|
|
taulukon osassa.
|
|
Segmenttipuun ylin luku on pienin luku
|
|
koko taulukon alueella.
|
|
Puun toteutus on samanlainen kuin summan laskemisessa,
|
|
mutta joka kohdassa pitää laskea summan sijasta
|
|
lukujen minimi.
|
|
|
|
\subsubsection{Binäärihaku puussa}
|
|
|
|
Segmenttipuun sisältämää tietoa voi käyttää
|
|
binäärihaun kaltaisesti aloittamalla
|
|
haun puun huipulta.
|
|
Näin on mahdollista selvittää esimerkiksi
|
|
minimisegmenttipuusta $O(\log n)$-ajassa,
|
|
missä kohdassa on taulukon pienin luku.
|
|
|
|
Esimerkiksi seuraavassa puussa pienin
|
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alkio on 1, jonka sijainti löytyy
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kulkemalla puussa huipulta alaspäin:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (8,0) grid (16,1);
|
|
|
|
\node[anchor=center] at (8.5, 0.5) {9};
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|
\node[anchor=center] at (9.5, 0.5) {5};
|
|
\node[anchor=center] at (10.5, 0.5) {7};
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|
\node[anchor=center] at (11.5, 0.5) {1};
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|
\node[anchor=center] at (12.5, 0.5) {6};
|
|
\node[anchor=center] at (13.5, 0.5) {2};
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|
\node[anchor=center] at (14.5, 0.5) {3};
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\node[anchor=center] at (15.5, 0.5) {2};
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|
|
|
%\node[anchor=center] at (1,2.5) {13};
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|
|
|
\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);
|
|
\path[draw,thick,-] (g) -- (13.5,1);
|
|
\node[draw, circle,minimum size=22pt] (h) at (15,2.5) {2};
|
|
\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{Lisätekniikoita}
|
|
|
|
\subsubsection{Indeksien pakkaus}
|
|
|
|
Taulukon päälle rakennettujen tietorakenteiden
|
|
rajoituksena on, että alkiot on indeksoitu
|
|
kokonaisluvuin $1,2,3,$ jne.
|
|
Tästä seuraa ongelmia,
|
|
jos tarvittavat indeksit ovat suuria.
|
|
Esimerkiksi indeksin $10^9$ käyttäminen
|
|
vaatisi, että taulukossa olisi $10^9$ alkiota,
|
|
mikä ei ole realistista.
|
|
|
|
\index{indeksien pakkaus@indeksien pakkaus}
|
|
|
|
Tätä rajoitusta on kuitenkin mahdollista
|
|
kiertää usein käyttämällä \key{indeksien pakkausta},
|
|
jolloin indeksit jaetaan
|
|
uudestaan niin, että ne ovat
|
|
kokonaisluvut $1,2,3,$ jne.
|
|
Tämä on mahdollista silloin, kun kaikki
|
|
algoritmin aikana tarvittavat indeksit
|
|
ovat tiedossa algoritmin alussa.
|
|
|
|
Ideana on korvata jokainen alkuperäinen
|
|
indeksi $x$ indeksillä $p(x)$,
|
|
missä $p$ jakaa indeksit uudestaan.
|
|
Vaatimuksena on, että indeksien järjestys
|
|
ei muutu, eli jos $a<b$, niin $p(a)<p(b)$,
|
|
minkä ansiosta kyselyitä voi tehdä
|
|
melko tavallisesti indeksien pakkauksesta huolimatta.
|
|
|
|
Esimerkiksi jos alkuperäiset indeksit ovat
|
|
$555$, $10^9$ ja $8$, ne muuttuvat näin:
|
|
|
|
\[
|
|
\begin{array}{lcl}
|
|
p(8) & = & 1 \\
|
|
p(555) & = & 2 \\
|
|
p(10^9) & = & 3 \\
|
|
\end{array}
|
|
\]
|
|
|
|
\subsubsection{Välin muuttaminen}
|
|
|
|
Tähän asti olemme toteuttaneet tietorakenteita,
|
|
joissa voi tehdä tehokkaasti välikyselyitä
|
|
ja muuttaa yksittäisiä taulukon arvoja.
|
|
Tarkastellaan lopuksi käänteistä tilannetta,
|
|
jossa pitääkin muuttaa välejä ja
|
|
kysellä yksittäisiä arvoja.
|
|
Keskitymme operaatioon,
|
|
joka kasvattaa kaikkia välin $[a,b]$ arvoja $x$:llä.
|
|
|
|
Yllättävää kyllä,
|
|
voimme käyttää tämän luvun tietorakenteita myös tässä tilanteessa.
|
|
Tämä vaatii, että muutamme taulukkoa niin,
|
|
että jokainen taulukon arvo kertoo \textit{muutoksen}
|
|
edelliseen arvoon nähden.
|
|
Esimerkiksi taulukosta
|
|
|
|
\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}
|
|
tulee seuraava:
|
|
\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}
|
|
|
|
Minkä tahansa vanhan arvon saa uudesta taulukosta
|
|
laskemalla summan taulukon alusta kyseiseen kohtaan asti.
|
|
Esimerkiksi kohdan 6 vanha arvo 5 saadaan
|
|
summana $3-2+4=5$.
|
|
|
|
Uuden tallennustavan etuna on,
|
|
että välin muuttamiseen riittää muuttaa
|
|
kahta taulukon kohtaa.
|
|
Esimerkiksi jos välille $2 \ldots 5$
|
|
lisätään luku 5,
|
|
taulukon kohtaan 2 lisätään 5
|
|
ja taulukon kohdasta 6 poistetaan 5.
|
|
Tulos on tässä:
|
|
|
|
\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}
|
|
|
|
Yleisemmin kun taulukon välille $a \ldots b$
|
|
lisätään $x$, taulukon kohtaan $a$
|
|
lisätään $x$ ja taulukon kohdasta $b+1$
|
|
vähennetään $x$.
|
|
Tarvittavat operaatiot
|
|
ovat summan laskeminen
|
|
taulukon alusta tiettyyn kohtaan
|
|
sekä yksittäisen alkion muuttaminen,
|
|
joten voimme käyttää tuttuja menetelmiä tässäkin tilanteessa.
|
|
|
|
Hankalampi tilanne on, jos samaan aikaan pitää pystyä
|
|
sekä kysymään tietoa väleiltä että muuttamaan välejä.
|
|
Myöhemmin luvussa 28 tulemme näkemään,
|
|
että tämäkin on mahdollista.
|
|
|
|
|
|
|