POJ 2155 解题报告

来源：互联网发布：gta5没网络怎么玩单机编辑：程序博客网时间：2024/05/19 17:57

这道题最简单的方法是用二维树状数组。

一维树状数组比较常见，资料也比较多：

TopCoder: https://www.topcoder.com/community/data-science/data-science-tutorials/binary-indexed-%20trees/

http://dongxicheng.org/structure/binary_indexed_tree/

看完后大致了解了。

但是这道题需要用到二维树状数组，并且是“区域更新，单点查询”，这也与常见用法不一致。所以我也不清楚为什么需要这样更新和查询（重点在于查询出来的不是区域么？），尽管博客讲解很详细：

http://www.cnblogs.com/chenxiwenruo/p/3388783.html

thestoryofsnow2155Accepted4088K422MSC++1520B

/* ID: thestor1 LANG: C++ TASK: poj2155 */#include <iostream>#include <fstream>#include <cmath>#include <cstdio>#include <cstring>#include <limits>#include <string>#include <vector>#include <list>#include <set>#include <map>#include <queue>#include <stack>#include <algorithm>#include <cassert>using namespace std;const int MAXN = 1000;int N, T;int C[MAXN + 1][MAXN + 1];inline int lowbit(int x){return x & (-x);}void updateBIT(int x, int y, int delta){for (int i = x; i > 0; i -= lowbit(i)){for (int j = y; j > 0; j -= lowbit(j)){C[i][j] += delta;}}}int queryBIT(int x, int y){int sum = 0;for (int i = x; i <= N; i += lowbit(i)){for (int j = y; j <= N; j += lowbit(j)){sum += C[i][j];}}return sum;}int main(){int X;scanf("%d", &X);char type;int x1, y1, x2, y2;for (int x = 0; x < X; ++x){scanf("%d%d", &N, &T);for (int i = 1; i <= N; ++i){for (int j = 1; j <= N; ++j){C[i][j] = 0;}}for (int t = 0; t < T; ++t){scanf(" %c ", &type);if (type == 'C'){scanf("%d%d%d%d", &x1, &y1, &x2, &y2);updateBIT(x2, y2, 1);updateBIT(x1 - 1, y2, -1);updateBIT(x2, y1 - 1, -1);updateBIT(x1 - 1, y1 - 1, 1);}else if (type == 'Q'){scanf("%d%d", &x1, &y1);printf("%d\n", queryBIT(x1, y1) % 2);}else{assert(false);}}// There is a blank line between every two continuous test cases. printf("\n");}return 0;  }

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