POJ2155 Matrix(树状数组)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total 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

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

Sample Output

1
0
0
1

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;
}