Improvements [closes #44]
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Antti H S Laaksonen
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chapter09.tex
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chapter09.tex
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@ -6,16 +6,17 @@
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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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that allow us to efficiently process range queries.
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In a \key{range query},
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our task is to calculate a value
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based on a subarray of an array.
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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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\item $\texttt{sum}_q(a,b)$: calculate the sum of values in range $[a,b]$
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\item $\texttt{min}_q(a,b)$: find the minimum value in range $[a,b]$
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\item $\texttt{max}_q(a,b)$: find the maximum value in range $[a,b]$
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\end{itemize}
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For example, consider the range $[3,6]$ in the following array:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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@ -42,38 +43,37 @@ For example, consider the range $[3,6]$ in the following array:
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\node at (7.5,1.4) {$7$};
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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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In this case, $\texttt{sum}_q(3,6)=14$,
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$\texttt{min}_q(3,6)=1$ and $\texttt{max}_q(3,6)=6$.
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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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A simple way to process range queries is to use
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a loop that goes through all array values in the range.
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For example, the following function can be
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used to process sum queries on an array:
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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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s += array[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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This function works in $O(n)$ time,
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where $n$ is the size of 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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is slow. Fortunately, 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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the array is \emph{static}, i.e.,
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the array values are never updated 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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@ -83,11 +83,12 @@ the answer for any possible query.
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\index{prefix sum array}
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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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sum queries on a static array
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by constructing a \key{prefix sum array}.
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Each value in the prefix sum array equals
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the sum of values in the original array up to that position,
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i.e., the value at position $k$ is $\texttt{sum}_q(0,k)$.
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The prefix sum array can be constructed in $O(n)$ time.
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For example, consider the following array:
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\begin{center}
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@ -142,14 +143,12 @@ The corresponding prefix sum array is as follows:
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\node at (7.5,1.4) {$7$};
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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 $\textrm{sum}(0,k)$,
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of $\texttt{sum}_q(0,k)$,
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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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\[ \textrm{sum}(a,b) = \textrm{sum}(0,b) - \textrm{sum}(0,a-1).\]
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By defining $\textrm{sum}(0,-1)=0$,
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$\texttt{sum}_q(a,b)$ in $O(1)$ time as follows:
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\[ \texttt{sum}_q(a,b) = \texttt{sum}_q(0,b) - \texttt{sum}_q(0,a-1)\]
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By defining $\texttt{sum}_q(0,-1)=0$,
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the above formula also holds when $a=0$.
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For example, consider the range $[3,6]$:
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@ -178,9 +177,9 @@ For example, consider the range $[3,6]$:
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\node at (7.5,1.4) {$7$};
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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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In this case $\texttt{sum}_q(3,6)=8+6+1+4=19$.
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This sum can be calculated from
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two values of 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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@ -196,7 +195,6 @@ two values in the prefix sum array:
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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) {$0$};
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\node at (1.5,1.4) {$1$};
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@ -208,14 +206,15 @@ two values in the prefix sum array:
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\node at (7.5,1.4) {$7$};
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\end{tikzpicture}
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\end{center}
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Thus, the sum in the range $[3,6]$ is $27-8=19$.
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Thus, $\texttt{sum}_q(3,6)=\texttt{sum}_q(0,6)-\texttt{sum}_q(0,2)=27-8=19$.
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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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prefix sum array that can be used to calculate
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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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Each sum in such an array corresponds to
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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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@ -224,7 +223,6 @@ The following picture illustrates the idea:
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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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@ -236,28 +234,33 @@ The following picture illustrates the idea:
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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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where $S(X)$ denotes the sum of values
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in 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 using \index{sparse table}
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a data structure called a \key{sparse table}\footnote{The
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sparse table structure was introduced in \cite{ben00}.
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There are also more sophisticated techniques \cite{fis06} where
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the preprocessing time of the array is only $O(n)$, but such algorithms
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are not needed in 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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\index{sparse table}
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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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Minimum queries are more difficult to process
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than sum queries.
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Still, there is a quite simple
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$O(n \log n)$ time preprocessing
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method after which we can answer any minimum
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query in $O(1)$ time\footnote{This technique
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was introduced in \cite{ben00} and sometimes
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called the \key{sparse table} method.
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There are also more sophisticated techniques \cite{fis06} where
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the preprocessing time is only $O(n)$, but such algorithms
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are not needed in competitive programming.}.
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Note that since minimum and maximum queries can
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be processed similarly,
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we can focus on minimum queries.
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The idea is to precalculate all values of
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$\textrm{min}_q(a,b)$ where
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$b-a+1$ (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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@ -284,13 +287,13 @@ For example, for the array
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\node at (7.5,1.4) {$7$};
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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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the following values are 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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\begin{tabular}{lll}
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$a$ & $b$ & $\texttt{min}_q(a,b)$ \\
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\hline
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0 & 0 & 1 \\
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1 & 1 & 3 \\
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@ -304,8 +307,8 @@ $a$ & $b$ & $\textrm{rmq}(a,b)$ \\
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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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\begin{tabular}{lll}
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$a$ & $b$ & $\texttt{min}_q(a,b)$ \\
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\hline
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0 & 1 & 1 \\
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1 & 2 & 3 \\
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@ -319,8 +322,8 @@ $a$ & $b$ & $\textrm{rmq}(a,b)$ \\
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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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\begin{tabular}{lll}
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$a$ & $b$ & $\texttt{min}_q(a,b)$ \\
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\hline
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0 & 3 & 1 \\
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1 & 4 & 3 \\
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@ -338,17 +341,17 @@ $a$ & $b$ & $\textrm{rmq}(a,b)$ \\
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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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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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\[\texttt{min}_q(a,b) = \min(\texttt{min}_q(a,a+w-1),\texttt{min}_q(a+w,b)),\]
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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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After this, any value of $\texttt{min}_q(a,b)$ 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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We can calculate the value of $\texttt{min}_q(a,b)$ using the formula
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\[\texttt{min}_q(a,b) = \min(\texttt{min}_q(a,a+k-1),\texttt{min}_q(b-k+1,b)).\]
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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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@ -434,41 +437,45 @@ the union of the ranges $[1,4]$ and $[3,6]$:
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\node at (7.5,1.4) {$7$};
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\end{tikzpicture}
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\end{center}
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Since $\textrm{rmq}(1,4)=3$ and $\textrm{rmq}(3,6)=1$,
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we can conclude that $\textrm{rmq}(1,6)=1$.
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Since $\texttt{min}_q(1,4)=3$ and $\texttt{min}_q(3,6)=1$,
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we conclude that $\texttt{min}_q(1,6)=1$.
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\section{Binary indexed trees}
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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}\footnote{The
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binary indexed tree structure was presented by P. M. Fenwick in 1994 \cite{fen94}.}
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can be seen as a dynamic version of a prefix sum array.
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This data structure supports two $O(\log n)$ time operations:
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calculating the sum of elements in a range
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and modifying the value of an element.
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can be seen as a dynamic variant of a prefix sum array.
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It supports two $O(\log n)$ time operations on an array:
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processing a range sum query and updating a value.
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The advantage of a binary indexed tree is
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that it allows us to efficiently \emph{update}
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array elements between sum queries.
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that it allows us to efficiently update
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array values between sum queries.
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This would not be possible using a prefix sum array,
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because after each update, it would be necessary to build the
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whole prefix sum array again in $O(n)$ time.
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\subsubsection{Structure}
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In this section we assume that one-based indexing
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is used, because it makes the implementation easier.
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A binary indexed tree is as an array
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whose value at position $x$
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equals the sum of elements in the range $[x-k+1,x]$
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of the original array,
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where $k$ is the largest power of two that divides $x$.
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For example, if $x=6$, then $k=2$, because 2 divides 6
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but 4 does not divide 6.
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Even if the name of the structure is a binary indexed \emph{tree},
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it is usually represented as an array.
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In this section we assume that all arrays are one-indexed,
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because it makes the implementation easier.
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Let $p(k)$ denote the largest power of two that
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divides $k$.
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We store a binary indexed tree as an array \texttt{tree}
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such that
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\[ \texttt{tree}[k] = \texttt{sum}_q(k-p(k)+1,k),\]
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i.e., each position $k$ contains the sum of values
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in a range of the original array whose length is $p(k)$
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and that ends at position $k$.
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For example, since $p(6)=2$, $\texttt{tree}[6]$
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contains the value of $\texttt{sum}_q(5,6)$.
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\begin{samepage}
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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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@ -494,8 +501,7 @@ For example, consider the following array:
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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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\end{samepage}
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\begin{samepage}
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The corresponding binary indexed tree is as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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@ -521,20 +527,13 @@ The corresponding binary indexed tree is as follows:
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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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\end{samepage}
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For example, the value at position 6
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in the binary indexed tree is 7,
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because the sum of elements in the range $[5,6]$
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of the array is $6+1=7$.
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The following picture shows more clearly
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how each value in the binary indexed tree
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corresponds to a range in the array:
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corresponds to a range in the original 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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@ -576,20 +575,16 @@ corresponds to a range in the array:
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\end{tikzpicture}
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\end{center}
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\subsubsection{Sum queries}
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Using a binary indexed tree,
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any value of $\texttt{sum}_q(1,k)$
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can be calculated in $O(\log n)$ time,
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because a range $[1,k]$ can always be divided into
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$O(\log n)$ ranges whose sums are stored in the tree.
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The values in a binary indexed tree
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can be used to efficiently calculate
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the sum of array elements in any range $[1,k]$,
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because such a range
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can be divided into $O(\log n)$ ranges
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whose sums are available in the binary indexed tree.
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For example, the range $[1,7]$ corresponds to
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the following values:
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For example, the range $[1,7]$ consists of
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the following ranges:
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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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@ -630,23 +625,23 @@ the following values:
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\draw (0,-4) -- (8,-4) -- (8,-4.5) -- (0,-4.5) -- (0,-4);
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\end{tikzpicture}
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\end{center}
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Thus, we can calculate the corresponding sum as follows:
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\[\texttt{sum}_q(1,7)=\texttt{sum}_q(1,4)+\texttt{sum}_q(5,6)+\texttt{sum}_q(7,7)=16+7+4=27\]
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Hence, the sum of elements in the range $[1,7]$ is $16+7+4=27$.
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To calculate the sum of elements in any range $[a,b]$,
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To calculate the value of $\texttt{sum}_q(a,b)$ where $a>1$,
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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.
|
||||
\[ \texttt{sum}_q(a,b) = \texttt{sum}_q(1,b) - \texttt{sum}_q(1,a-1).\]
|
||||
Since we can calculate both $\texttt{sum}_q(1,b)$
|
||||
and $\texttt{sum}_q(1,a-1)$ in $O(\log n)$ time,
|
||||
the total time complexity is $O(\log n)$.
|
||||
|
||||
\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,
|
||||
Then, after updating a value in the original array,
|
||||
several values in the binary indexed tree
|
||||
should be updated.
|
||||
For example, if the value 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) {$1$};
|
||||
|
|
@ -690,45 +685,35 @@ the sums of the following ranges change:
|
|||
|
||||
Since each array element belongs to $O(\log n)$
|
||||
ranges in the binary indexed tree,
|
||||
it suffices to update $O(\log n)$ values.
|
||||
|
||||
it suffices to update $O(\log n)$ values in the tree.
|
||||
|
||||
\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.
|
||||
The operations of a binary indexed tree can be
|
||||
efficiently implemented using bit operations.
|
||||
The key fact needed is that we can
|
||||
calculate any value of $p(k)$ using the formula
|
||||
\[p(k) = k \& -k.\]
|
||||
|
||||
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 $[1,k]$:
|
||||
The following function calculates the value of $\texttt{sum}_q(1,k)$:
|
||||
\begin{lstlisting}
|
||||
int sum(int k) {
|
||||
int s = 0;
|
||||
while (k >= 1) {
|
||||
s += b[k];
|
||||
s += tree[k];
|
||||
k -= k&-k;
|
||||
}
|
||||
return s;
|
||||
}
|
||||
\end{lstlisting}
|
||||
|
||||
The following function increases the value
|
||||
of the element at position $k$ by $x$
|
||||
The following function increases the
|
||||
array value 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;
|
||||
tree[k] += x;
|
||||
k += k&-k;
|
||||
}
|
||||
}
|
||||
|
|
@ -737,20 +722,18 @@ void add(int k, int x) {
|
|||
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.
|
||||
to the next position takes $O(1)$ time.
|
||||
|
||||
\section{Segment trees}
|
||||
\section{Segment tree}
|
||||
|
||||
\index{segment tree}
|
||||
|
||||
A \key{segment tree}\footnote{Quite similar structures were used
|
||||
in late 1970's to solve geometric problems \cite{ben80}.
|
||||
The bottom-up-implementation in this chapter corresponds to
|
||||
that in \cite{sta06}.} is a data structure
|
||||
A \key{segment tree}\footnote{The bottom-up-implementation in this chapter corresponds to
|
||||
that in \cite{sta06}. Similar structures were used
|
||||
in late 1970's to solve geometric problems \cite{ben80}.} is a data structure
|
||||
that supports two operations:
|
||||
processing a range query and
|
||||
modifying an element in the array.
|
||||
updating an array value.
|
||||
Segment trees can support
|
||||
sum queries, minimum and maximum queries and many other
|
||||
queries so that both operations work in $O(\log n)$ time.
|
||||
|
|
@ -774,7 +757,7 @@ correspond to the array elements,
|
|||
and the other nodes
|
||||
contain information needed for processing range queries.
|
||||
|
||||
Throughout the section, we assume that the size
|
||||
In this 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.
|
||||
|
|
@ -847,19 +830,18 @@ The corresponding segment tree is as follows:
|
|||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Each internal node in the segment tree contains
|
||||
information about a range of size $2^k$
|
||||
in the original array.
|
||||
Each internal tree node
|
||||
corresponds to an array range
|
||||
whose size is a power of two.
|
||||
In the above tree, the value of each internal
|
||||
node is the sum of the corresponding array elements,
|
||||
node is the sum of the corresponding array values,
|
||||
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:
|
||||
It turns out that any range $[a,b]$
|
||||
can be divided into $O(\log n)$ ranges
|
||||
whose values are stored in tree nodes.
|
||||
For example, consider the range [2,7]:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.7]
|
||||
\fill[color=gray!50] (2,0) rectangle (8,1);
|
||||
|
|
@ -873,21 +855,20 @@ For example, consider the following range:
|
|||
\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) {$8$};
|
||||
|
||||
\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 sum of elements in the range is
|
||||
$6+3+2+7+2+6=26$.
|
||||
The following two nodes in the tree
|
||||
Here $\texttt{sum}_q(2,7)=6+3+2+7+2+6=26$.
|
||||
In this case, the following two tree nodes
|
||||
correspond to the range:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.7]
|
||||
|
|
@ -927,27 +908,24 @@ correspond to the range:
|
|||
\path[draw,thick,-] (m) -- (j);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
Thus, the sum of elements in the range is $9+17=26$.
|
||||
Thus, another way to calculate the sum is $9+17=26$.
|
||||
|
||||
When the sum is calculated using nodes
|
||||
that are located as high as possible in the tree,
|
||||
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 $O(\log n)$.
|
||||
is $O(\log n)$.
|
||||
|
||||
\subsubsection{Array updates}
|
||||
|
||||
When an element in the array changes,
|
||||
we should update all nodes in the tree
|
||||
whose value depends on the element.
|
||||
After an array update,
|
||||
we should update all nodes
|
||||
whose value depends on the updated value.
|
||||
This can be done by traversing the path
|
||||
from the element to the top node
|
||||
from the updated array 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.
|
||||
The following picture shows which tree nodes
|
||||
change if the array value 7 changes:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.7]
|
||||
|
|
@ -988,27 +966,24 @@ change if the element 7 in the array changes.
|
|||
\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}
|
||||
\subsubsection{Implementation}
|
||||
|
||||
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.
|
||||
We store a segment tree as an array
|
||||
of $2n$ elements where $n$ is the size of
|
||||
the original array and a power of two.
|
||||
The tree nodes are stored from top to bottom:
|
||||
$\texttt{tree}[1]$ is the top node,
|
||||
$\texttt{tree}[2]$ and $\texttt{tree}[3]$
|
||||
are its children, and so on.
|
||||
Finally, the values from $\texttt{tree}[n]$
|
||||
to $\texttt{tree}[2n-1]$ correspond to
|
||||
the values of the original array
|
||||
on the bottom level of the tree.
|
||||
|
||||
For example, the segment tree
|
||||
\begin{center}
|
||||
|
|
@ -1049,10 +1024,9 @@ For example, the segment tree
|
|||
\path[draw,thick,-] (m) -- (j);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
can be stored as follows ($N=8$):
|
||||
is stored as follows:
|
||||
\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) {$39$};
|
||||
|
|
@ -1090,79 +1064,67 @@ can be stored as follows ($N=8$):
|
|||
\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.
|
||||
the parent of $\texttt{tree}[k]$
|
||||
is $\texttt{tree}[\lfloor k/2 \rfloor]$,
|
||||
and its children are $\texttt{tree}[2k]$
|
||||
and $\texttt{tree}[2k+1]$.
|
||||
Note that this implies that the position 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 $[a,b]$:
|
||||
|
||||
calculates the value of $\texttt{sum}_q(a,b)$:
|
||||
\begin{lstlisting}
|
||||
int sum(int a, int b) {
|
||||
a += N; b += N;
|
||||
a += n; b += n;
|
||||
int s = 0;
|
||||
while (a <= b) {
|
||||
if (a%2 == 1) s += p[a++];
|
||||
if (b%2 == 0) s += p[b--];
|
||||
if (a%2 == 1) s += tree[a++];
|
||||
if (b%2 == 0) s += tree[b--];
|
||||
a /= 2; b /= 2;
|
||||
}
|
||||
return s;
|
||||
}
|
||||
\end{lstlisting}
|
||||
The function maintains a range
|
||||
that is initially $[a+n,b+n]$.
|
||||
Then, at each step, the range is moved
|
||||
one level higher in the tree,
|
||||
and before that, the values of the nodes that do not
|
||||
belong to the higher range are added to the sum.
|
||||
|
||||
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 $x$:
|
||||
|
||||
The following function increases the array value
|
||||
at position $k$ by $x$:
|
||||
\begin{lstlisting}
|
||||
void add(int k, int x) {
|
||||
k += N;
|
||||
p[k] += x;
|
||||
k += n;
|
||||
tree[k] += x;
|
||||
for (k /= 2; k >= 1; k /= 2) {
|
||||
p[k] = p[2*k]+p[2*k+1];
|
||||
tree[k] = tree[2*k]+tree[2*k+1];
|
||||
}
|
||||
}
|
||||
\end{lstlisting}
|
||||
First the function updates the element
|
||||
First the function updates the value
|
||||
at the bottom level of the tree.
|
||||
After this, the function updates the values of all
|
||||
internal nodes in the tree, until it reaches
|
||||
internal tree nodes, until it reaches
|
||||
the top node of the tree.
|
||||
|
||||
Both above functions work
|
||||
Both the 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.
|
||||
and the functions move one level higher
|
||||
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,
|
||||
Segment trees can support all range queries
|
||||
where it is possible to divide a range into two parts,
|
||||
calculate the answer separately 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:
|
||||
|
||||
|
|
@ -1204,71 +1166,65 @@ supports minimum queries:
|
|||
\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.
|
||||
In this case, every tree node contains
|
||||
the smallest value in the corresponding
|
||||
array range.
|
||||
The top node of the tree contains the smallest
|
||||
element of the whole array.
|
||||
value 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.
|
||||
The structure of a segment tree also allows us
|
||||
to use binary search for locating array elements.
|
||||
For example, if the tree supports minimum queries,
|
||||
we can find the position of the smallest
|
||||
element in $O(\log n)$ time.
|
||||
we can find the position of an element
|
||||
with the smallest value in $O(\log n)$ time.
|
||||
|
||||
For example, in the following tree the
|
||||
smallest element 1 can be found
|
||||
For example, in the above tree, an
|
||||
element with the smallest value 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);
|
||||
\draw (0,0) grid (8,1);
|
||||
|
||||
\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 (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[anchor=center] at (1,2.5) {13};
|
||||
\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] (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] (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] (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] (m) at (4,6.5) {1};
|
||||
\path[draw,thick,-] (m) -- (i);
|
||||
\path[draw,thick,-] (m) -- (j);
|
||||
|
||||
\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);
|
||||
\path[draw=red,thick,->,line width=2pt] (m) -- (j);
|
||||
\path[draw=red,thick,->,line width=2pt] (j) -- (c);
|
||||
\path[draw=red,thick,->,line width=2pt] (c) -- (4.5,1);
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
|
@ -1296,11 +1252,11 @@ 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
|
||||
with $c(x)$ where $c$ 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
|
||||
does not change, so if $a<b$, then $c(a)<c(b)$.
|
||||
This allows us to conveniently perform queries
|
||||
even if the indices are compressed.
|
||||
|
||||
For example, if the original indices are
|
||||
|
|
@ -1308,9 +1264,9 @@ $555$, $10^9$ and $8$, the new indices are:
|
|||
|
||||
\[
|
||||
\begin{array}{lcl}
|
||||
p(8) & = & 1 \\
|
||||
p(555) & = & 2 \\
|
||||
p(10^9) & = & 3 \\
|
||||
c(8) & = & 1 \\
|
||||
c(555) & = & 2 \\
|
||||
c(10^9) & = & 3 \\
|
||||
\end{array}
|
||||
\]
|
||||
|
||||
|
|
@ -1330,9 +1286,8 @@ elements in a range $[a,b]$ by $x$.
|
|||
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
|
||||
whose values indicate
|
||||
the differences between consecutive values
|
||||
in the original array.
|
||||
Thus, the original array is the
|
||||
prefix sum array of the
|
||||
|
|
@ -1393,15 +1348,17 @@ The difference array for the above array is as follows:
|
|||
\end{center}
|
||||
|
||||
For example, the value 2 at position 6 in the original array
|
||||
corresponds to the sum $3-2+4-3=2$.
|
||||
corresponds to the sum $3-2+4-3=2$ in the difference array.
|
||||
|
||||
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 $1 \ldots 4$ by 5,
|
||||
it suffices to increase the value at position 1 by 5
|
||||
increase the original array
|
||||
values between positions 1 and 4 by 5,
|
||||
it suffices to increase the
|
||||
difference array value at position 1 by 5
|
||||
and decrease the value at position 5 by 5.
|
||||
The result is as follows:
|
||||
|
||||
|
|
@ -1430,8 +1387,8 @@ The result is as follows:
|
|||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
More generally, to increase the elements
|
||||
in a range $[a,b]$ by $x$,
|
||||
More generally, to increase the values
|
||||
in 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
|
||||
|
|
|
|||
Loading…
Reference in New Issue