POJ 2155 Matrix

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y]. 

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above. 

Output

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

Sample Input

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

给一个n*n矩阵 初始化值全为0。 然后有m个指示进行两个操作
操作1：给出子矩阵的左上角和右下角坐标，矩阵里的元素 1变0 ，0 变1

操作2：给出询问，问当前点(x,y)的值是多少。

这个写得好
http://blog.csdn.net/zxy_snow/article/details/6264135

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <stack>#include <string>#include <map>#include <set>#define pi acos(-1)#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef pair<int, int> P;const int maxn = 1e3 + 5;int n, m;int C[maxn<<2][maxn<<2];int lowbit(int x){return x & (-x); }void update(int x, int y){int i, j;for (i = x; i <= n; i += lowbit(i))for (j = y; j <= n; j += lowbit(j))C[i][j]++;}int Getsum(int x, int y){int res = 0, i, j;for (i = x; i != 0; i -= lowbit(i))for (j = y; j != 0; j -= lowbit(j))res += C[i][j];return res;}int main(void){//freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin);int T, i, j, x1, y1, x2, y2;char s[2];scanf("%d", &T);while (T--){memset(C, 0, sizeof(C));scanf("%d %d", &n, &m);while(m--){scanf("%s", &s);if (s[0] == 'C'){scanf("%d %d %d %d", &x1, &y1, &x2, &y2);update(x1, y1);update(x1, y2+1);update(x2+1, y1);update(x2+1, y2+1);}else {scanf("%d %d", &x1, &y1);printf("%d\n", Getsum(x1, y1) % 2);}} printf("\n");}return 0;}