POJMatrix（二维树状数组）

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 22058		Accepted: 8219

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，跟一维的差不多，
这个道题的思路就是看看x1,y1往上加一，同时方块右边，下面和右下方的区域再加1，只要保证他们那边加个偶数就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 1000 + 5;
int c[MAX][MAX];
int n;
int lowbit(int k)
{
    return k & (-k);
}
void update(int x,int y,int num)
{
    for(int i = x; i < n; i += lowbit(i))
    {
        for(int j = y; j < n; j += lowbit(j))
            c[i][j] += num;
    }
}
int sum(int x,int y)
{
    int s = 0;
    for(int i = x; i > 0; i -= lowbit(i))
    {
        for(int j = y; j > 0; j -= lowbit(j))
            s += c[i][j];
    }
    return s;
}
int main()
{
    int t,q;
    int num = 0;
    scanf("%d", &t);
    while(t--)
    {
        if(num ++)
            printf("
");
        scanf("%d%d", &n,&q);
        memset(c,0,sizeof(c));
        char ch;
        int x1,y1,x2,y2;
        getchar();
        while(q--)
        {
            scanf("%c", &ch);
            if(ch == 'Q')
            {
                scanf("%d%d", &x1,&y1);
                getchar();
                int m = sum(x2,y2) - sum(x1 - 1, y2) - sum(x2,y1 - 1) + sum(x1-1,y1-1);
                if(m % 2 == 0)
                    printf("0
");
                else
                    printf("1
");
            }
            else if(ch == 'C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                getchar();
                update(x1,y1,1);
            }
        }
    }
    return 0;
}