POJ2155 Matrix(树状数组)

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 21448 Accepted: 8016

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

二维树状数组的应用，注意t不为0时输出一个空行。

AC代码：

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"using namespace std;const int MAXN = 1100;int a[MAXN][MAXN], n;int lowbit(int x){return x & (-x);}void update(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))a[i][j]++;}int get_sum(int x, int y){int ans = 0;for(int i = x; i > 0; i -= lowbit(i))for(int j = y; j > 0; j -= lowbit(j))ans += a[i][j];return ans;}int main(int argc, char const *argv[]){int t, m;scanf("%d", &t);while(t--) {memset(a, 0, sizeof(a));scanf("%d%d", &n, &m);while(m--) {char op[2];scanf("%s", op);if(op[0] == 'C') {int c, x1, x2, y1, y2;scanf("%d%d%d%d", &x1, &y1, &x2, &y2);x1++, y1++, x2++, y2++;update(x2, y2);update(x1 - 1, y1 - 1);update(x1 - 1, y2);update(x2, y1 - 1);}else {int x, y;scanf("%d%d", &x, &y);printf("%d\n", get_sum(x, y) % 2);}}if(t != 0) printf("\n");}return 0;}