Matrix_二维树状数组

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

【题意】给出一个n*n的方阵，刚开始都为0；然后q个要求，c x1 y1 x2,y2表示左上角（x1,y1）右下角（x2,y2）的这个矩形里面的各方格与现在状态相反（即0变1,1变0），Q x1 y1问（x1,y1）的状态。

【思路】二维树状数组，每次状态变化都加一，最后对2取余

参考资料：http://www.cnblogs.com/lvpengms/archive/2010/04/24/1719133.html

当对(x1,y1),(x2,y2)区间置反时，需要改动四个地方就是4个角就可以了。为什么呢？如下图，假设A区未需要置反的区域，因为改动A区的左上角时，由树状数组的性质知：A,B,C,D4个区域都是要被置反的，所以在依次置反BD,CD,D，这样，置反的总过程为ABCD,BD,CD,D，这样我们就会发现结果对2取模时，只有A区被置反，B,C,D三个区都没有变化。明白原理之后就好做了。

A	B
C	D

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int N=1005;
int c[N][N];
int q,n;
int lowbit(int x)
{
    return x&(-x);
}
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+=c[i][j];
    return ans;
}
void update(int x,int y,int data)
{
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=n;j+=lowbit(j))
        c[i][j]+=data;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&q);
        memset(c,0,sizeof(c));
        char op[10];
        int x1,y1,x2,y2;
        while(q--)
        {
            scanf("%s",op);
            if(op[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                update(x1,y1,1);
                update(x2+1,y1,1);
                update(x1,y2+1,1);
                update(x2+1,y2+1,1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                if(get_sum(x1,y1)%2==0) puts("0");
                else puts("1");
            }
        }
        if(t>0) puts("");
    }
    return 0;
}