POJ2155 Matrix（二维树状数组||区间修改单点查询）

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：

在一个N*N的矩阵里（左上是(1,1)），初始点的值都为0，C(x1,y1,x2,y2)表示将这个矩阵的点都异或，Q(x,y)表示查询点的值（0或者1）。

思路：

常见的二维树状数组是单点更新，区间查询； 而这里是区间更新，单点查询。

由于是单点查询，这里直接用差分的思想做的：a[i][j]表示坐标(i,j)到(n,m)增加多少。

如果矩形(x1,y1,x2,y2)加一，则a[x1][x2]+1;a[x1][y2+1]-1;a[x2][y1+1]-1,a[x2][y2]+1;那么所求点(i,j)的值就是前缀和。

（但如果是区间更新，区间查询，则要像上一题那样推公式，最后得到5个一维树状数组。上一题是3个一维树状数组。）

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m,a[][];
int lowbit(int x){return x&(-x);}
void add(int x,int y,int val)
{
    for(int i=x;i<=n;i+=lowbit(i))
     for(int j=y;j<=n;j+=lowbit(j))
      a[i][j]+=val;
}
int query(int x,int y)
{
    int res=;
    for(int i=x;i;i-=lowbit(i))
     for(int j=y;j;j-=lowbit(j))
      res+=a[i][j];
    return res;
}
int main()
{
    int T,x1,x2,y1,y2;char opt[];scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        memset(a,,sizeof(a));
        for(int i=;i<=m;i++){
            scanf("%s",opt);
            if(opt[]=='Q'){
                scanf("%d%d",&x1,&y1);
                printf("%d\n",&(query(x1,y1)));
            }
            else {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x1,y1,);add(x2+,y2+,);
                add(x2+,y1,-);add(x1,y2+,-);
            }
        }
        if(T) printf("\n");
    } return ;
}