POJ 1195 解题报告

来源：互联网发布：万捷网络验证系统编辑：程序博客网时间：2024/05/18 00:38

这道题是二维树状数组（BIT）的基本应用。之前已经写过2155，那道题是区域更新，单点输出，比较异常。这道题是基本的单点更新，区域输出，所以比较好理解，一次就过了。

thestoryofsnow1195Accepted4276K594MSC++

/* 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 <= N; i += lowbit(i)){for (int j = y; j <= N; j += lowbit(j)){C[i][j] += delta;}}}int queryBIT(int x, int y){int sum = 0;for (int i = x; i > 0; i -= lowbit(i)){for (int j = y; j > 0; 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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