1169 lines
37 KiB
TeX
1169 lines
37 KiB
TeX
\chapter{Segment trees revisited}
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\index{segment tree}
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A segment tree is a versatile data structure
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that can be used in many different situations.
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However, there are many topics related to segment trees
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that we have not touched yet.
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Now it is time to discuss some more advanced variants
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of segment trees.
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So far, we have implemented the operations
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of a segment tree by walking \emph{from bottom to top}
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in the tree.
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For example, we have calculated the sum of
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elements in a range $[a,b]$
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as follows (Chapter 9.3):
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\begin{lstlisting}
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int sum(int a, int b) {
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a += N; b += N;
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int s = 0;
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while (a <= b) {
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if (a%2 == 1) s += p[a++];
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if (b%2 == 0) s += p[b--];
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a /= 2; b /= 2;
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}
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return s;
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}
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\end{lstlisting}
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However, in more advanced segment trees,
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it is often needed to implement the operations
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in another way, \emph{from top to bottom}.
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Using this approach, the function becomes as follows:
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\begin{lstlisting}
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int sum(int a, int b, int k, int x, int y) {
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if (b < x || a > y) return 0;
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if (a == x && b == y) return p[k];
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int d = (y-x+1)/2;
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return sum(a, min(x+d-1,b), 2*k, x, x+d-1) +
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sum(max(x+d,a), b, 2*k+1, x+d, y);
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}
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\end{lstlisting}
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Now we can calculate the sum of
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elements in $[a,b]$ as follows:
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\begin{lstlisting}
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int s = sum(a, b, 1, 0, N-1);
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\end{lstlisting}
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\begin{samepage}
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The parameter $k$ indicates the current position
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in \texttt{p}.
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Initially $k$ equals 1, because we begin
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at the root of the segment tree.
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The range $[x,y]$ corresponds to $k$
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and is initially $[0,N-1]$.
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When calculating the sum,
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if $[a,b]$ is outside $[x,y]$,
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the sum is 0,
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and if $[a,b]$ equals $[x,y]$,
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the sum can be found in \texttt{p}.
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If $[a,b]$ is completely or partially inside $[x,y]$,
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the search continues recursively to the
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left and right half of $[x,y]$.
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The size of both halves is $d=\frac{1}{2}(y-x+1)$;
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the left half is $[x,x+d-1]$
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and the right half is $[x+d,y]$.
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\end{samepage}
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The following picture shows how the search proceeds
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when calculating the sum of elements in $[a,b]$.
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The gray nodes indicate nodes where the recursion
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stops and the sum can be found in \texttt{p}.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=gray!50] (5,0) rectangle (6,1);
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\draw (0,0) grid (16,1);
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\node[anchor=center] at (0.5, 0.5) {5};
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\node[anchor=center] at (1.5, 0.5) {8};
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\node[anchor=center] at (2.5, 0.5) {6};
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\node[anchor=center] at (3.5, 0.5) {3};
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\node[anchor=center] at (4.5, 0.5) {2};
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\node[anchor=center] at (5.5, 0.5) {7};
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\node[anchor=center] at (6.5, 0.5) {2};
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\node[anchor=center] at (7.5, 0.5) {6};
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\node[anchor=center] at (8.5, 0.5) {7};
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\node[anchor=center] at (9.5, 0.5) {1};
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\node[anchor=center] at (10.5, 0.5) {7};
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\node[anchor=center] at (11.5, 0.5) {5};
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\node[anchor=center] at (12.5, 0.5) {6};
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\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] (a) at (1,2.5) {13};
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\path[draw,thick,-] (a) -- (0.5,1);
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\path[draw,thick,-] (a) -- (1.5,1);
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\node[draw, circle,minimum size=22pt] (b) at (3,2.5) {9};
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\path[draw,thick,-] (b) -- (2.5,1);
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\path[draw,thick,-] (b) -- (3.5,1);
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\node[draw, circle,minimum size=22pt] (c) at (5,2.5) {9};
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\path[draw,thick,-] (c) -- (4.5,1);
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\path[draw,thick,-] (c) -- (5.5,1);
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\node[draw, circle,fill=gray!50,minimum size=22pt] (d) at (7,2.5) {8};
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\path[draw,thick,-] (d) -- (6.5,1);
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\path[draw,thick,-] (d) -- (7.5,1);
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\node[draw, circle,minimum size=22pt] (e) at (9,2.5) {8};
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\path[draw,thick,-] (e) -- (8.5,1);
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\path[draw,thick,-] (e) -- (9.5,1);
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\node[draw, circle] (f) at (11,2.5) {12};
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\path[draw,thick,-] (f) -- (10.5,1);
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\path[draw,thick,-] (f) -- (11.5,1);
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\node[draw, circle,fill=gray!50,minimum size=22pt] (g) at (13,2.5) {8};
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\path[draw,thick,-] (g) -- (12.5,1);
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\path[draw,thick,-] (g) -- (13.5,1);
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\node[draw, circle,minimum size=22pt] (h) at (15,2.5) {5};
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\path[draw,thick,-] (h) -- (14.5,1);
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\path[draw,thick,-] (h) -- (15.5,1);
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\node[draw, circle] (i) at (2,4.5) {22};
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\path[draw,thick,-] (i) -- (a);
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\path[draw,thick,-] (i) -- (b);
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\node[draw, circle] (j) at (6,4.5) {17};
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\path[draw,thick,-] (j) -- (c);
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\path[draw,thick,-] (j) -- (d);
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\node[draw, circle,fill=gray!50] (k) at (10,4.5) {20};
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\path[draw,thick,-] (k) -- (e);
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\path[draw,thick,-] (k) -- (f);
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\node[draw, circle] (l) at (14,4.5) {13};
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\path[draw,thick,-] (l) -- (g);
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\path[draw,thick,-] (l) -- (h);
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\node[draw, circle] (m) at (4,6.5) {39};
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\path[draw,thick,-] (m) -- (i);
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\path[draw,thick,-] (m) -- (j);
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\node[draw, circle] (n) at (12,6.5) {33};
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\path[draw,thick,-] (n) -- (k);
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\path[draw,thick,-] (n) -- (l);
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\node[draw, circle] (o) at (8,8.5) {72};
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\path[draw,thick,-] (o) -- (m);
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\path[draw,thick,-] (o) -- (n);
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\path[draw=red,thick,->,line width=2pt] (o) -- (m);
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\path[draw=red,thick,->,line width=2pt] (o) -- (n);
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\path[draw=red,thick,->,line width=2pt] (m) -- (j);
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\path[draw=red,thick,->,line width=2pt] (j) -- (c);
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\path[draw=red,thick,->,line width=2pt] (j) -- (d);
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\path[draw=red,thick,->,line width=2pt] (c) -- (5.5,1);
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\path[draw=red,thick,->,line width=2pt] (n) -- (k);
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\path[draw=red,thick,->,line width=2pt] (n) -- (l);
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\path[draw=red,thick,->,line width=2pt] (l) -- (g);
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\draw [decoration={brace}, decorate, line width=0.5mm] (14,-0.25) -- (5,-0.25);
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\node at (5.5,-0.75) {$a$};
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\node at (13.5,-0.75) {$b$};
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\end{tikzpicture}
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\end{center}
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Also in this implementation,
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operations take $O(\log n)$ time,
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because the total number of visited nodes is $O(\log n)$.
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\section{Lazy propagation}
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\index{lazy propagation}
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\index{lazy segment tree}
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Using \key{lazy propagation}, we can build
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a segment tree that supports both range updates
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and range queries in $O(\log n)$ time.
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The idea is to perform updates and queries
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from top to bottom and perform updates
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\emph{lazily} so that they are propagated
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down the tree only when it is necessary.
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In a lazy segment tree, nodes contain two types of
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information.
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Like in an ordinary segment tree,
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each node contains the sum or some other value
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related to the corresponding subarray.
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In addition, the node may contain information
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related to lazy updates, which has not been
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propagated to its children.
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There are two possible types of range updates:
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each element in the range is either
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\emph{increased} by some value
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or \emph{assigned} some value.
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Both operations can be implemented using
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similar ideas, and it is even possible to construct
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a tree that supports both operations at the same time.
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\subsubsection{Lazy segment tree}
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Let us consider an example where our goal is to
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construct a segment tree that supports
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two operations: increasing each element in
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$[a,b]$ by $u$ and calculating the sum of
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elements in $[a,b]$.
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We will construct a tree where each node
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contains two values $s/z$:
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$s$ denotes the sum of elements in the range
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and $z$ denotes the value of a lazy update,
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which means that all elements in the range
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should be increased by $z$.
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In the following tree, $z=0$ for all nodes,
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so there are no ongoing lazy updates.
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (0,0) grid (16,1);
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\node[anchor=center] at (0.5, 0.5) {5};
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\node[anchor=center] at (1.5, 0.5) {8};
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\node[anchor=center] at (2.5, 0.5) {6};
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\node[anchor=center] at (3.5, 0.5) {3};
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\node[anchor=center] at (4.5, 0.5) {2};
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\node[anchor=center] at (5.5, 0.5) {7};
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\node[anchor=center] at (6.5, 0.5) {2};
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\node[anchor=center] at (7.5, 0.5) {6};
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\node[anchor=center] at (8.5, 0.5) {7};
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\node[anchor=center] at (9.5, 0.5) {1};
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\node[anchor=center] at (10.5, 0.5) {7};
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\node[anchor=center] at (11.5, 0.5) {5};
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\node[anchor=center] at (12.5, 0.5) {6};
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\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[draw, circle] (a) at (1,2.5) {13/0};
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\path[draw,thick,-] (a) -- (0.5,1);
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\path[draw,thick,-] (a) -- (1.5,1);
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\node[draw, circle,minimum size=32pt] (b) at (3,2.5) {9/0};
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\path[draw,thick,-] (b) -- (2.5,1);
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\path[draw,thick,-] (b) -- (3.5,1);
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\node[draw, circle,minimum size=32pt] (c) at (5,2.5) {9/0};
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\path[draw,thick,-] (c) -- (4.5,1);
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\path[draw,thick,-] (c) -- (5.5,1);
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\node[draw, circle,minimum size=32pt] (d) at (7,2.5) {8/0};
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\path[draw,thick,-] (d) -- (6.5,1);
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\path[draw,thick,-] (d) -- (7.5,1);
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\node[draw, circle,minimum size=32pt] (e) at (9,2.5) {8/0};
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\path[draw,thick,-] (e) -- (8.5,1);
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\path[draw,thick,-] (e) -- (9.5,1);
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\node[draw, circle] (f) at (11,2.5) {12/0};
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\path[draw,thick,-] (f) -- (10.5,1);
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\path[draw,thick,-] (f) -- (11.5,1);
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\node[draw, circle,minimum size=32pt] (g) at (13,2.5) {8/0};
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\path[draw,thick,-] (g) -- (12.5,1);
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\path[draw,thick,-] (g) -- (13.5,1);
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\node[draw, circle,minimum size=32pt] (h) at (15,2.5) {5/0};
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\path[draw,thick,-] (h) -- (14.5,1);
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\path[draw,thick,-] (h) -- (15.5,1);
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\node[draw, circle] (i) at (2,4.5) {22/0};
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\path[draw,thick,-] (i) -- (a);
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\path[draw,thick,-] (i) -- (b);
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\node[draw, circle] (j) at (6,4.5) {17/0};
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\path[draw,thick,-] (j) -- (c);
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\path[draw,thick,-] (j) -- (d);
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\node[draw, circle] (k) at (10,4.5) {20/0};
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\path[draw,thick,-] (k) -- (e);
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\path[draw,thick,-] (k) -- (f);
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\node[draw, circle] (l) at (14,4.5) {13/0};
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\path[draw,thick,-] (l) -- (g);
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\path[draw,thick,-] (l) -- (h);
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\node[draw, circle] (m) at (4,6.5) {39/0};
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\path[draw,thick,-] (m) -- (i);
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\path[draw,thick,-] (m) -- (j);
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\node[draw, circle] (n) at (12,6.5) {33/0};
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\path[draw,thick,-] (n) -- (k);
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\path[draw,thick,-] (n) -- (l);
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\node[draw, circle] (o) at (8,8.5) {72/0};
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\path[draw,thick,-] (o) -- (m);
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\path[draw,thick,-] (o) -- (n);
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\end{tikzpicture}
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\end{center}
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When the elements in $[a,b]$ are increased by $u$,
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we walk from the root towards the leaves
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and modify the nodes in the tree as follows:
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If the range $[x,y]$ of a node is
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completely inside the range $[a,b]$,
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we increase the $z$ value of the node by $u$ and stop.
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If $[x,y]$ only partially belongs to $[a,b]$,
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we increase the $s$ value of the node by $hu$,
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where $h$ is the size of the intersection of $[a,b]$
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and $[x,y]$, and continue our walk recursively in the tree.
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For example, the following picture shows the tree after
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increasing the elements in the range $[a,b]$ by 2:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
|
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\fill[color=gray!50] (5,0) rectangle (6,1);
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\draw (0,0) grid (16,1);
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|
|
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\node[anchor=center] at (0.5, 0.5) {5};
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\node[anchor=center] at (1.5, 0.5) {8};
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\node[anchor=center] at (2.5, 0.5) {6};
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\node[anchor=center] at (3.5, 0.5) {3};
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\node[anchor=center] at (4.5, 0.5) {2};
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\node[anchor=center] at (5.5, 0.5) {9};
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\node[anchor=center] at (6.5, 0.5) {2};
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\node[anchor=center] at (7.5, 0.5) {6};
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\node[anchor=center] at (8.5, 0.5) {7};
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\node[anchor=center] at (9.5, 0.5) {1};
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\node[anchor=center] at (10.5, 0.5) {7};
|
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\node[anchor=center] at (11.5, 0.5) {5};
|
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\node[anchor=center] at (12.5, 0.5) {6};
|
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\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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|
|
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\node[draw, circle] (a) at (1,2.5) {13/0};
|
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\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
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\node[draw, circle,minimum size=32pt] (b) at (3,2.5) {9/0};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
\node[draw, circle,minimum size=32pt] (c) at (5,2.5) {11/0};
|
|
\path[draw,thick,-] (c) -- (4.5,1);
|
|
\path[draw,thick,-] (c) -- (5.5,1);
|
|
\node[draw, circle,fill=gray!50,minimum size=32pt] (d) at (7,2.5) {8/2};
|
|
\path[draw,thick,-] (d) -- (6.5,1);
|
|
\path[draw,thick,-] (d) -- (7.5,1);
|
|
\node[draw, circle,minimum size=32pt] (e) at (9,2.5) {8/0};
|
|
\path[draw,thick,-] (e) -- (8.5,1);
|
|
\path[draw,thick,-] (e) -- (9.5,1);
|
|
\node[draw, circle] (f) at (11,2.5) {12/0};
|
|
\path[draw,thick,-] (f) -- (10.5,1);
|
|
\path[draw,thick,-] (f) -- (11.5,1);
|
|
\node[draw, circle,fill=gray!50,minimum size=32pt] (g) at (13,2.5) {8/2};
|
|
\path[draw,thick,-] (g) -- (12.5,1);
|
|
\path[draw,thick,-] (g) -- (13.5,1);
|
|
\node[draw, circle,minimum size=32pt] (h) at (15,2.5) {5/0};
|
|
\path[draw,thick,-] (h) -- (14.5,1);
|
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\path[draw,thick,-] (h) -- (15.5,1);
|
|
|
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\node[draw, circle] (i) at (2,4.5) {22/0};
|
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\path[draw,thick,-] (i) -- (a);
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\path[draw,thick,-] (i) -- (b);
|
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\node[draw, circle] (j) at (6,4.5) {23/0};
|
|
\path[draw,thick,-] (j) -- (c);
|
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\path[draw,thick,-] (j) -- (d);
|
|
\node[draw, circle,fill=gray!50] (k) at (10,4.5) {20/2};
|
|
\path[draw,thick,-] (k) -- (e);
|
|
\path[draw,thick,-] (k) -- (f);
|
|
\node[draw, circle] (l) at (14,4.5) {17/0};
|
|
\path[draw,thick,-] (l) -- (g);
|
|
\path[draw,thick,-] (l) -- (h);
|
|
|
|
\node[draw, circle] (m) at (4,6.5) {45/0};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\node[draw, circle] (n) at (12,6.5) {45/0};
|
|
\path[draw,thick,-] (n) -- (k);
|
|
\path[draw,thick,-] (n) -- (l);
|
|
|
|
\node[draw, circle] (o) at (8,8.5) {90/0};
|
|
\path[draw,thick,-] (o) -- (m);
|
|
\path[draw,thick,-] (o) -- (n);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (o) -- (m);
|
|
\path[draw=red,thick,->,line width=2pt] (o) -- (n);
|
|
|
|
\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] (j) -- (d);
|
|
\path[draw=red,thick,->,line width=2pt] (c) -- (5.5,1);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (n) -- (k);
|
|
\path[draw=red,thick,->,line width=2pt] (n) -- (l);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (l) -- (g);
|
|
|
|
\draw [decoration={brace}, decorate, line width=0.5mm] (14,-0.25) -- (5,-0.25);
|
|
|
|
\node at (5.5,-0.75) {$a$};
|
|
\node at (13.5,-0.75) {$b$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
We also calculate the sum of elements in a range $[a,b]$
|
|
by walking in the tree from top to bottom.
|
|
If the range $[x,y]$ of a node completely belongs
|
|
to $[a,b]$, we add the $s$ value of the node to the sum.
|
|
Otherwise, we continue the search recursively
|
|
downwards in the tree.
|
|
|
|
Both in updates and queries,
|
|
the value of a lazy update is always propagated
|
|
to the children of the node
|
|
before processing the node.
|
|
The idea is that updates will be propagated
|
|
downwards only when it is necessary,
|
|
which guarantees that the operations are always efficient.
|
|
|
|
The following picture shows how the tree changes
|
|
when we calculate the sum of elements in $[a,b]$.
|
|
The rectangle shows the nodes whose values change,
|
|
because a lazy update is propagated downwards,
|
|
which is necessary to calculate the sum in $[a,b]$.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (16,1);
|
|
|
|
\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) {9};
|
|
\node[anchor=center] at (6.5, 0.5) {2};
|
|
\node[anchor=center] at (7.5, 0.5) {6};
|
|
\node[anchor=center] at (8.5, 0.5) {7};
|
|
\node[anchor=center] at (9.5, 0.5) {1};
|
|
\node[anchor=center] at (10.5, 0.5) {7};
|
|
\node[anchor=center] at (11.5, 0.5) {5};
|
|
\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[draw, circle] (a) at (1,2.5) {13/0};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,minimum size=32pt] (b) at (3,2.5) {9/0};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
\node[draw, circle,minimum size=32pt] (c) at (5,2.5) {11/0};
|
|
\path[draw,thick,-] (c) -- (4.5,1);
|
|
\path[draw,thick,-] (c) -- (5.5,1);
|
|
\node[draw, circle,minimum size=32pt] (d) at (7,2.5) {8/2};
|
|
\path[draw,thick,-] (d) -- (6.5,1);
|
|
\path[draw,thick,-] (d) -- (7.5,1);
|
|
\node[draw, circle,minimum size=32pt] (e) at (9,2.5) {8/2};
|
|
\path[draw,thick,-] (e) -- (8.5,1);
|
|
\path[draw,thick,-] (e) -- (9.5,1);
|
|
\node[draw, circle,fill=gray!50,] (f) at (11,2.5) {12/2};
|
|
\path[draw,thick,-] (f) -- (10.5,1);
|
|
\path[draw,thick,-] (f) -- (11.5,1);
|
|
\node[draw, circle,fill=gray!50,minimum size=32pt] (g) at (13,2.5) {8/2};
|
|
\path[draw,thick,-] (g) -- (12.5,1);
|
|
\path[draw,thick,-] (g) -- (13.5,1);
|
|
\node[draw, circle,minimum size=32pt] (h) at (15,2.5) {5/0};
|
|
\path[draw,thick,-] (h) -- (14.5,1);
|
|
\path[draw,thick,-] (h) -- (15.5,1);
|
|
|
|
\node[draw, circle] (i) at (2,4.5) {22/0};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\node[draw, circle] (j) at (6,4.5) {23/0};
|
|
\path[draw,thick,-] (j) -- (c);
|
|
\path[draw,thick,-] (j) -- (d);
|
|
\node[draw, circle] (k) at (10,4.5) {28/0};
|
|
\path[draw,thick,-] (k) -- (e);
|
|
\path[draw,thick,-] (k) -- (f);
|
|
\node[draw, circle] (l) at (14,4.5) {17/0};
|
|
\path[draw,thick,-] (l) -- (g);
|
|
\path[draw,thick,-] (l) -- (h);
|
|
|
|
\node[draw, circle] (m) at (4,6.5) {45/0};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\node[draw, circle] (n) at (12,6.5) {45/0};
|
|
\path[draw,thick,-] (n) -- (k);
|
|
\path[draw,thick,-] (n) -- (l);
|
|
|
|
\node[draw, circle] (o) at (8,8.5) {90/0};
|
|
\path[draw,thick,-] (o) -- (m);
|
|
\path[draw,thick,-] (o) -- (n);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (o) -- (n);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (n) -- (k);
|
|
\path[draw=red,thick,->,line width=2pt] (n) -- (l);
|
|
|
|
\path[draw=red,thick,->,line width=2pt] (k) -- (f);
|
|
\path[draw=red,thick,->,line width=2pt] (l) -- (g);
|
|
|
|
\draw [decoration={brace}, decorate, line width=0.5mm] (14,-0.25) -- (10,-0.25);
|
|
|
|
\draw[color=blue,thick] (8,1.5) rectangle (12,5.5);
|
|
|
|
\node at (10.5,-0.75) {$a$};
|
|
\node at (13.5,-0.75) {$b$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Note that sometimes it is needed to combine lazy updates.
|
|
This happens when a node that already has a lazy update
|
|
is assigned another lazy update.
|
|
In this problem, it is easy to combine lazy updates,
|
|
because the combination of updates $z_1$ and $z_2$
|
|
corresponds to an update $z_1+z_2$.
|
|
|
|
\subsubsection{Polynomial update}
|
|
|
|
Lazy updates can be generalized so that it is
|
|
possible to update ranges using polynomials of the form
|
|
\[p(u) = t_k u^k + t_{k-1} u^{k-1} + \cdots + t_0.\]
|
|
|
|
In this case, the update for an element $i$
|
|
in the range $[a,b]$ is $p(i-a)$.
|
|
For example, adding $p(u)=u+1$
|
|
to $[a,b]$ means that the element at position $a$
|
|
is increased by 1, the element at position $a+1$
|
|
is increased by 2, etc.
|
|
|
|
To support polynomial updates,
|
|
each node is assigned $k+2$ values,
|
|
where $k$ equals the degree of the polynomial.
|
|
The value $s$ is the sum of the elements in the range,
|
|
and the values $z_0,z_1,\ldots,z_k$ are the coefficients
|
|
of a polynomial that corresponds to a lazy update.
|
|
|
|
Now, the sum of elements in a range $[x,y]$ equals
|
|
\[s+\sum_{u=0}^{y-x} z_k u^k + z_{k-1} u^{k-1} + \cdots + z_0.\]
|
|
|
|
The value of such a sum
|
|
can be efficiently calculated using sum formulas.
|
|
For example, the term $z_0$ corresponds to the sum
|
|
$(y-x+1)z_0$, and the term $z_1 u$ corresponds to the sum
|
|
\[z_1(0+1+\cdots+y-x) = z_1 \frac{(y-x)(y-x+1)}{2} .\]
|
|
|
|
When propagating an update in the tree,
|
|
the indices of $p(u)$ change,
|
|
because in each range $[x,y]$,
|
|
the values are
|
|
calculated for $u=0,1,\ldots,y-x$.
|
|
However, this is not a problem, because
|
|
$p'(u)=p(u+h)$ is a polynomial
|
|
of equal degree as $p(u)$.
|
|
For example, if $p(u)=t_2 u^2+t_1 u-t_0$, then
|
|
\[p'(u)=t_2(u+h)^2+t_1(u+h)-t_0=t_2 u^2 + (2ht_2+t_1)u+t_2h^2+t_1h-t_0.\]
|
|
|
|
\section{Dynamic trees}
|
|
|
|
\index{dynamic segment tree}
|
|
|
|
An ordinary segment tree is static,
|
|
which means that each node has a fixed position
|
|
in the array and the tree requires
|
|
a fixed amount of memory.
|
|
However, if most nodes are not used,
|
|
memory is wasted.
|
|
In a \key{dynamic segment tree},
|
|
memory is allocated only for nodes that
|
|
are actually needed.
|
|
|
|
The nodes of a dynamic tree can be represented as structs:
|
|
|
|
\begin{lstlisting}
|
|
struct node {
|
|
int s;
|
|
int x, y;
|
|
node *l, *r;
|
|
node(int x, int a, int b) : s(s), x(x), y(y) {}
|
|
};
|
|
\end{lstlisting}
|
|
Here $s$ is the value of the node,
|
|
$[x,y]$ is the corresponding range,
|
|
and $l$ and $r$ point to the left
|
|
and right subtree.
|
|
|
|
After this, nodes can be created as follows:
|
|
|
|
\begin{lstlisting}
|
|
// create new node
|
|
node *u = new node(0, 0, 15);
|
|
// change value
|
|
u->s = 5;
|
|
\end{lstlisting}
|
|
|
|
\subsubsection{Sparse segment tree}
|
|
|
|
\index{sparse segment tree}
|
|
|
|
A dynamic segment tree is useful if
|
|
the underlying array is \emph{sparse}.
|
|
This means that the range $[0,N-1]$
|
|
of allowed indices is large,
|
|
but only a small portion of the indices are used
|
|
and most elements in the array are empty.
|
|
While an ordinary segment tree uses $O(N)$ memory,
|
|
a dynamic segment tree only requires $O(n \log N)$ memory,
|
|
where $n$ is the number of operations performed.
|
|
|
|
A \key{sparse segment tree} is initially empty
|
|
and its only node is $[0,N-1]$.
|
|
After updates, new nodes are added dynamically
|
|
when needed.
|
|
For example, if $N=16$ and the elements
|
|
in positions 3 and 10 have been modified,
|
|
the tree contains the following nodes:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\scriptsize
|
|
\node[draw, circle,minimum size=35pt] (1) at (0,0) {$[0,15]$};
|
|
\node[draw, circle,minimum size=35pt] (2) at (-4,-2) {$[0,7]$};
|
|
\node[draw, circle,minimum size=35pt] (3) at (-6,-4) {$[0,3]$};
|
|
\node[draw, circle,minimum size=35pt] (4) at (-4,-6) {$[2,3]$};
|
|
\node[draw, circle,minimum size=35pt] (5) at (-2,-8) {$[3]$};
|
|
\node[draw, circle,minimum size=35pt] (6) at (4,-2) {$[8,15]$};
|
|
\node[draw, circle,minimum size=35pt] (7) at (2,-4) {$[8,11]$};
|
|
\node[draw, circle,minimum size=35pt] (8) at (4,-6) {$[10,11]$};
|
|
\node[draw, circle,minimum size=35pt] (9) at (2,-8) {$[10]$};
|
|
|
|
\path[draw,thick,->] (1) -- (2);
|
|
\path[draw,thick,->] (2) -- (3);
|
|
\path[draw,thick,->] (3) -- (4);
|
|
\path[draw,thick,->] (4) -- (5);
|
|
|
|
\path[draw,thick,->] (1) -- (6);
|
|
\path[draw,thick,->] (6) -- (7);
|
|
\path[draw,thick,->] (7) -- (8);
|
|
\path[draw,thick,->] (8) -- (9);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Any path from the root node to a leaf contains
|
|
$O(\log N)$ nodes,
|
|
so each operation adds at most $O(\log N)$
|
|
new nodes to the tree.
|
|
Thus, after $n$ operations, the tree contains
|
|
at most $O(n \log N)$ nodes.
|
|
|
|
Note that if all indices of the elements
|
|
are known at the beginning of the algorithm,
|
|
a dynamic segment tree is not needed,
|
|
but we can use an ordinary segment tree with
|
|
index compression (Chapter 9.4).
|
|
However, this is not possible if the indices
|
|
are generated during the algorithm.
|
|
|
|
\subsubsection{Persistent segment tree}
|
|
|
|
\index{persistent segment tree}
|
|
|
|
Using a dynamic implementation,
|
|
it is also possible to create a
|
|
\key{persistent segment tree} that stores
|
|
the \key{modification history} of the tree.
|
|
In such an implementation, we can
|
|
efficiently access
|
|
all versions of the tree that have
|
|
existed during the algorithm.
|
|
|
|
When the modification history is available,
|
|
we can perform queries in any previous tree
|
|
like in an ordinary segment tree, because the
|
|
full structure of each tree is stored.
|
|
We can also create new trees based on previous
|
|
trees and modify them independently.
|
|
|
|
Consider the following sequence of updates,
|
|
where red nodes change
|
|
and other nodes remain the same:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.8]
|
|
\node[draw, circle,minimum size=13pt] (1a) at (3,0) {};
|
|
\node[draw, circle,minimum size=13pt] (2a) at (2,-1) {};
|
|
\node[draw, circle,minimum size=13pt] (3a) at (4,-1) {};
|
|
\node[draw, circle,minimum size=13pt] (4a) at (1.5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (5a) at (2.5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (6a) at (3.5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (7a) at (4.5,-2) {};
|
|
\path[draw,thick,->] (1a) -- (2a);
|
|
\path[draw,thick,->] (1a) -- (3a);
|
|
\path[draw,thick,->] (2a) -- (4a);
|
|
\path[draw,thick,->] (2a) -- (5a);
|
|
\path[draw,thick,->] (3a) -- (6a);
|
|
\path[draw,thick,->] (3a) -- (7a);
|
|
|
|
\node[draw, circle,minimum size=13pt,fill=red] (1b) at (3+5,0) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (2b) at (2+5,-1) {};
|
|
\node[draw, circle,minimum size=13pt] (3b) at (4+5,-1) {};
|
|
\node[draw, circle,minimum size=13pt] (4b) at (1.5+5,-2) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (5b) at (2.5+5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (6b) at (3.5+5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (7b) at (4.5+5,-2) {};
|
|
\path[draw,thick,->] (1b) -- (2b);
|
|
\path[draw,thick,->] (1b) -- (3b);
|
|
\path[draw,thick,->] (2b) -- (4b);
|
|
\path[draw,thick,->] (2b) -- (5b);
|
|
\path[draw,thick,->] (3b) -- (6b);
|
|
\path[draw,thick,->] (3b) -- (7b);
|
|
|
|
\node[draw, circle,minimum size=13pt,fill=red] (1c) at (3+10,0) {};
|
|
\node[draw, circle,minimum size=13pt] (2c) at (2+10,-1) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (3c) at (4+10,-1) {};
|
|
\node[draw, circle,minimum size=13pt] (4c) at (1.5+10,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (5c) at (2.5+10,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (6c) at (3.5+10,-2) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (7c) at (4.5+10,-2) {};
|
|
\path[draw,thick,->] (1c) -- (2c);
|
|
\path[draw,thick,->] (1c) -- (3c);
|
|
\path[draw,thick,->] (2c) -- (4c);
|
|
\path[draw,thick,->] (2c) -- (5c);
|
|
\path[draw,thick,->] (3c) -- (6c);
|
|
\path[draw,thick,->] (3c) -- (7c);
|
|
|
|
\node at (3,-3) {step 1};
|
|
\node at (3+5,-3) {step 2};
|
|
\node at (3+10,-3) {step 3};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
After each update, most nodes in the tree
|
|
remain the same,
|
|
so an efficient way to store the modification history
|
|
is to represent each tree in the history as a combination
|
|
of new nodes and subtrees of previous trees.
|
|
In this example, the modification history can be
|
|
stored as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.8]
|
|
\path[use as bounding box] (0, 1) rectangle (16, -3.5);
|
|
|
|
\node[draw, circle,minimum size=13pt] (1a) at (3,0) {};
|
|
\node[draw, circle,minimum size=13pt] (2a) at (2,-1) {};
|
|
\node[draw, circle,minimum size=13pt] (3a) at (4,-1) {};
|
|
\node[draw, circle,minimum size=13pt] (4a) at (1.5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (5a) at (2.5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (6a) at (3.5,-2) {};
|
|
\node[draw, circle,minimum size=13pt] (7a) at (4.5,-2) {};
|
|
\path[draw,thick,->] (1a) -- (2a);
|
|
\path[draw,thick,->] (1a) -- (3a);
|
|
\path[draw,thick,->] (2a) -- (4a);
|
|
\path[draw,thick,->] (2a) -- (5a);
|
|
\path[draw,thick,->] (3a) -- (6a);
|
|
\path[draw,thick,->] (3a) -- (7a);
|
|
|
|
\node[draw, circle,minimum size=13pt,fill=red] (1b) at (3+5,0) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (2b) at (2+5,-1) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (5b) at (2.5+5,-2) {};
|
|
\path[draw,thick,->] (1b) -- (2b);
|
|
|
|
\draw[thick,->] (1b) .. controls (3+5+2,0-1) and (3+5,2.5) .. (3a);
|
|
|
|
\draw[thick,->] (2b) .. controls (2+5-0.5,-1-0.5) and (2,4.5) .. (4a);
|
|
|
|
|
|
\path[draw,thick,->] (2b) -- (5b);
|
|
|
|
\node[draw, circle,minimum size=13pt,fill=red] (1c) at (3+10,0) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (3c) at (4+10,-1) {};
|
|
\node[draw, circle,minimum size=13pt,fill=red] (7c) at (4.5+10,-2) {};
|
|
\path[draw,thick,->] (1c) -- (2b);
|
|
\path[draw,thick,->] (1c) -- (3c);
|
|
|
|
\draw[thick,->] (3c) .. controls (2.5+5,-3) and (3.5,-3) .. (6a);
|
|
|
|
\path[draw,thick,->] (3c) -- (7c);
|
|
|
|
\node at (3,-3) {step 1};
|
|
\node at (3+5,-3) {step 2};
|
|
\node at (3+10,-3) {step 3};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The structure of each previous tree can be
|
|
reconstructed by following the pointers
|
|
starting at the corresponding root node.
|
|
Since each operation during the algorithm
|
|
adds only $O(\log N)$ new nodes to the tree,
|
|
it is possible to store the full modification history of the tree.
|
|
|
|
\section{Data structures}
|
|
|
|
Instead of single values, nodes in a segment tree
|
|
can also contain data structures that maintain information
|
|
about the corresponding ranges.
|
|
In such a tree, the operations take
|
|
$O(f(n) \log n)$ time, where $f(n)$ is
|
|
the time needed for processing a single node during an operation.
|
|
|
|
As an example, consider a segment tree that
|
|
supports queries of the form
|
|
''how many times does an element $x$ appear
|
|
in the range $[a,b]$?''
|
|
For example, the element 1 appears three times
|
|
in the following range:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\fill[lightgray] (1,0) rectangle (6,1);
|
|
\draw (0,0) grid (8,1);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {3};
|
|
\node[anchor=center] at (1.5, 0.5) {1};
|
|
\node[anchor=center] at (2.5, 0.5) {2};
|
|
\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) {1};
|
|
\node[anchor=center] at (6.5, 0.5) {1};
|
|
\node[anchor=center] at (7.5, 0.5) {2};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The idea is to build a segment tree
|
|
where each node is assigned a data structure
|
|
that can calculate the number of any element $x$
|
|
in the corresponding range.
|
|
Using such a segment tree,
|
|
the answer to a query can be calculated
|
|
by combining the results from the nodes
|
|
that belong to the range.
|
|
|
|
For example, the following segment tree
|
|
corresponds to the above array:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
|
|
\node[draw, rectangle] (a) at (1,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
3 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
\node[draw, rectangle] (b) at (3,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
1 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
\node[draw, rectangle] (c) at (5,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
2 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
\node[draw, rectangle] (d) at (7,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
3 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
\node[draw, rectangle] (e) at (9,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
1 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
\node[draw, rectangle] (f) at (11,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
1 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
\node[draw, rectangle] (g) at (13,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
1 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
\node[draw, rectangle] (h) at (15,2.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
2 \\
|
|
\hline
|
|
1 \\
|
|
\end{tabular}};
|
|
|
|
\node[draw, rectangle] (i) at (2,4.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{rr}
|
|
1 & 3 \\
|
|
\hline
|
|
1 & 1 \\
|
|
\end{tabular}};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\node[draw, rectangle] (j) at (6,4.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{rr}
|
|
2 & 3 \\
|
|
\hline
|
|
1 & 1 \\
|
|
\end{tabular}};
|
|
\path[draw,thick,-] (j) -- (c);
|
|
\path[draw,thick,-] (j) -- (d);
|
|
\node[draw, rectangle] (k) at (10,4.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{r}
|
|
1 \\
|
|
\hline
|
|
2 \\
|
|
\end{tabular}};
|
|
\path[draw,thick,-] (k) -- (e);
|
|
\path[draw,thick,-] (k) -- (f);
|
|
\node[draw, rectangle] (l) at (14,4.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{rr}
|
|
1 & 2 \\
|
|
\hline
|
|
1 & 1 \\
|
|
\end{tabular}};
|
|
\path[draw,thick,-] (l) -- (g);
|
|
\path[draw,thick,-] (l) -- (h);
|
|
|
|
\node[draw, rectangle] (m) at (4,6.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{rrr}
|
|
1 & 2 & 3 \\
|
|
\hline
|
|
1 & 1 & 2 \\
|
|
\end{tabular}};
|
|
\path[draw,thick,-] (m) -- (i);
|
|
\path[draw,thick,-] (m) -- (j);
|
|
\node[draw, rectangle] (n) at (12,6.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{rr}
|
|
1 & 2 \\
|
|
\hline
|
|
3 & 1 \\
|
|
\end{tabular}};
|
|
\path[draw,thick,-] (n) -- (k);
|
|
\path[draw,thick,-] (n) -- (l);
|
|
|
|
\node[draw, rectangle] (o) at (8,8.5)
|
|
{
|
|
\footnotesize
|
|
\begin{tabular}{rrr}
|
|
1 & 2 & 3 \\
|
|
\hline
|
|
4 & 2 & 2 \\
|
|
\end{tabular}};
|
|
\path[draw,thick,-] (o) -- (m);
|
|
\path[draw,thick,-] (o) -- (n);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
For example, we can build the tree so
|
|
that each node contains a \texttt{map} structure.
|
|
In this case, the time needed for processing each
|
|
node is $O(\log n)$, so the total time complexity
|
|
of a query is $O(\log^2 n)$.
|
|
The tree uses $O(n \log n)$ memory,
|
|
because there are $O(\log n)$ levels
|
|
and each level contains
|
|
$O(n)$ elements.
|
|
|
|
\section{Two-dimensionality}
|
|
|
|
\index{two-dimensional segment tree}
|
|
|
|
A \key{two-dimensional segment tree} supports
|
|
queries related to rectangular subarrays
|
|
of a two-dimensional array.
|
|
Such a tree can be implemented as
|
|
nested segment trees: a big tree corresponds to the
|
|
rows in the array, and each node contains a small tree
|
|
that corresponds to a column.
|
|
|
|
For example, in the array
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\draw (0,0) grid (4,4);
|
|
|
|
\node[anchor=center] at (0.5, 0.5) {8};
|
|
\node[anchor=center] at (1.5, 0.5) {5};
|
|
\node[anchor=center] at (2.5, 0.5) {3};
|
|
\node[anchor=center] at (3.5, 0.5) {8};
|
|
|
|
\node[anchor=center] at (0.5, 1.5) {3};
|
|
\node[anchor=center] at (1.5, 1.5) {9};
|
|
\node[anchor=center] at (2.5, 1.5) {7};
|
|
\node[anchor=center] at (3.5, 1.5) {1};
|
|
|
|
\node[anchor=center] at (0.5, 2.5) {8};
|
|
\node[anchor=center] at (1.5, 2.5) {7};
|
|
\node[anchor=center] at (2.5, 2.5) {5};
|
|
\node[anchor=center] at (3.5, 2.5) {2};
|
|
|
|
\node[anchor=center] at (0.5, 3.5) {7};
|
|
\node[anchor=center] at (1.5, 3.5) {6};
|
|
\node[anchor=center] at (2.5, 3.5) {1};
|
|
\node[anchor=center] at (3.5, 3.5) {6};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
the sum of any subarray
|
|
can be calculated
|
|
from the following segment tree:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.4]
|
|
\footnotesize
|
|
\begin{scope}[shift={(-12,0)}]
|
|
\draw (-1,-1) rectangle (5,6);
|
|
\draw (0,0) grid (4,1);
|
|
\node[anchor=center,scale=0.8] at (0.5, 0.5) {7};
|
|
\node[anchor=center,scale=0.8] at (1.5, 0.5) {6};
|
|
\node[anchor=center,scale=0.8] at (2.5, 0.5) {1};
|
|
\node[anchor=center,scale=0.8] at (3.5, 0.5) {6};
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {13};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,scale=0.8,inner sep=2.5pt] (b) at (3,2.5) {7};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {20};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\end{scope}
|
|
\begin{scope}[shift={(-4,0)}]
|
|
\draw (-1,-1) rectangle (5,6);
|
|
\draw (0,0) grid (4,1);
|
|
\node[anchor=center,scale=0.8] at (0.5, 0.5) {8};
|
|
\node[anchor=center,scale=0.8] at (1.5, 0.5) {7};
|
|
\node[anchor=center,scale=0.8] at (2.5, 0.5) {5};
|
|
\node[anchor=center,scale=0.8] at (3.5, 0.5) {2};
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {15};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,scale=0.8,inner sep=2.5pt] (b) at (3,2.5) {7};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {22};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\end{scope}
|
|
\begin{scope}[shift={(4,0)}]
|
|
\draw (-1,-1) rectangle (5,6);
|
|
\draw (0,0) grid (4,1);
|
|
\node[anchor=center,scale=0.8] at (0.5, 0.5) {3};
|
|
\node[anchor=center,scale=0.8] at (1.5, 0.5) {9};
|
|
\node[anchor=center,scale=0.8] at (2.5, 0.5) {7};
|
|
\node[anchor=center,scale=0.8] at (3.5, 0.5) {1};
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {12};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,scale=0.8,inner sep=2.5pt] (b) at (3,2.5) {8};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {20};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\end{scope}
|
|
\begin{scope}[shift={(12,0)}]
|
|
\draw (-1,-1) rectangle (5,6);
|
|
\draw (0,0) grid (4,1);
|
|
\node[anchor=center,scale=0.8] at (0.5, 0.5) {8};
|
|
\node[anchor=center,scale=0.8] at (1.5, 0.5) {5};
|
|
\node[anchor=center,scale=0.8] at (2.5, 0.5) {3};
|
|
\node[anchor=center,scale=0.8] at (3.5, 0.5) {8};
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {13};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {11};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {24};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\end{scope}
|
|
\begin{scope}[shift={(-8,10)}]
|
|
\draw (-1,-1) rectangle (5,6);
|
|
\draw (0,0) grid (4,1);
|
|
\node[anchor=center,scale=0.8] at (0.5, 0.5) {15};
|
|
\node[anchor=center,scale=0.8] at (1.5, 0.5) {13};
|
|
\node[anchor=center,scale=0.8] at (2.5, 0.5) {6};
|
|
\node[anchor=center,scale=0.8] at (3.5, 0.5) {8};
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {28};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {14};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {42};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\end{scope}
|
|
\begin{scope}[shift={(8,10)}]
|
|
\draw (-1,-1) rectangle (5,6);
|
|
\draw (0,0) grid (4,1);
|
|
\node[anchor=center,scale=0.8] at (0.5, 0.5) {11};
|
|
\node[anchor=center,scale=0.8] at (1.5, 0.5) {14};
|
|
\node[anchor=center,scale=0.8] at (2.5, 0.5) {10};
|
|
\node[anchor=center,scale=0.8] at (3.5, 0.5) {9};
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {25};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {19};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {44};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\end{scope}
|
|
\begin{scope}[shift={(0,20)}]
|
|
\draw (-1,-1) rectangle (5,6);
|
|
\draw (0,0) grid (4,1);
|
|
\node[anchor=center,scale=0.8] at (0.5, 0.5) {26};
|
|
\node[anchor=center,scale=0.8] at (1.5, 0.5) {27};
|
|
\node[anchor=center,scale=0.8] at (2.5, 0.5) {16};
|
|
\node[anchor=center,scale=0.8] at (3.5, 0.5) {17};
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (a) at (1,2.5) {53};
|
|
\path[draw,thick,-] (a) -- (0.5,1);
|
|
\path[draw,thick,-] (a) -- (1.5,1);
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (b) at (3,2.5) {33};
|
|
\path[draw,thick,-] (b) -- (2.5,1);
|
|
\path[draw,thick,-] (b) -- (3.5,1);
|
|
|
|
\node[draw, circle,scale=0.8,inner sep=1pt] (i) at (2,4.5) {86};
|
|
\path[draw,thick,-] (i) -- (a);
|
|
\path[draw,thick,-] (i) -- (b);
|
|
\end{scope}
|
|
\path[draw,thick,-] (2,19) -- (-6,16);
|
|
\path[draw,thick,-] (2,19) -- (10,16);
|
|
\path[draw,thick,-] (-6,9) -- (-10,6);
|
|
\path[draw,thick,-] (-6,9) -- (-2,6);
|
|
\path[draw,thick,-] (10,9) -- (6,6);
|
|
\path[draw,thick,-] (10,9) -- (14,6);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The operations in a two-dimensional segment tree
|
|
take $O(\log^2 n)$ time, because the big tree
|
|
and each small tree consist of $O(\log n)$ levels.
|
|
The tree requires $O(n^2)$ memory, because each
|
|
small tree contains $O(n)$ values.
|