poj-2155 Matrix(二维树状数组)

题目链接：

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 23170		Accepted: 8613

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
AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1e3+4;
int sum[N][N],n,l,r,xl,xr,yl,yr,m;
char c;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y,int num)
{

    while(y<=n)
    {
        int fx=x;
        while(fx<=n)
        {
            sum[fx][y]+=num;
            fx+=lowbit(fx);
        }
        y+=lowbit(y);
    }
}
int query(int x,int y)
{
    int s=0;
    while(y>0){
            int fx=x;
    while(fx>0)
    {
        s+=sum[fx][y];
        fx-=lowbit(fx);
    }
    y-=lowbit(y);
    }
    return s;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
            cin>>c;
            if(c=='C')
            {
                scanf("%d%d%d%d",&xl,&yl,&xr,&yr);
                update(xl,yl,1);
                update(xl,yr+1,1);
                update(xr+1,yl,1);
                update(xr+1,yr+1,1);
            }
            else
            {
                scanf("%d%d",&l,&r);
                printf("%d\n",query(l,r)%2);
            }
        }
        printf("\n");
    }
    return 0;
}