poj 2155 Matrix (树状数组)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16797		Accepted: 6312

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

Source

POJ Monthly,Lou Tiancheng

//4100K    438MS    C++    1052B    2014-04-01 21:00:26
/*

    题意：
        给出一个n*n的 二维区间，有小矩形取反操作和对某元素求值操作，对每个操作
    做出处理。
    
    树状数组：
        这题和一般的树状数组不一样，一般的是对某个值更改，对某段求值，现在是刚好
    反过来，利用反向思维，讲原先更新的操作用来求值， 求值的操作用来更新即可。 

*/
#include<stdio.h>
#include<string.h>
#define N 1005
int c[N][N];
int lowbit(int i)
{
    return i&(-i);
}
int update(int x,int y)
{
    int s=0;
    for(int i=x;i<N;i+=lowbit(i))
        for(int j=y;j<N;j+=lowbit(j))
            s+=c[i][j];
    return s%2;
            
}
void getsum(int x,int y)
{
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j))
            c[i][j]^=1;
}
int main(void)
{
    int t,n,m;
    char op;
    int x1,x2,y1,y2;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%*c",&n,&m);
        memset(c,0,sizeof(c));
        while(m--){
            scanf("%c",&op);
            if(op=='Q'){
                scanf("%d%d%*c",&x1,&y1);
                printf("%d
",update(x1,y1));
            }else{
                scanf("%d%d%d%d%*c",&x1,&y1,&x2,&y2);
                getsum(x1-1,y1-1);
                getsum(x1-1,y2);
                getsum(x2,y1-1);
                getsum(x2,y2);
            }
        }
        printf("
");
    }
    return 0;
}