poj 2155 Matrix---树状数组套树状数组

二维树状数组模版，唯一困难，看题！！（其实是我英语渣）
Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 22098 Accepted: 8240

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].
   题目大意：
   给出t个n*n的矩阵，初始都是0，并给一个m，给出m个命令：
   命令“C x1 y1 x2 y2”将（x1，y1）–（x2，y2）上每个点进行交换（0变为1，1变为0）
   命令“Q x y”求（x，y）的值
   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
前排膜娄教主%%%

这个题的话，值得一提的就是：
在修改(x1,y1)--(x2,y2)的时候，应用区间修改的原理只需要修改(x1,y1),(x2+1,y1),(x1,y2+1),(x2+1,y2+1)即可
以及每个记录的值是变换过几次的值，所以结果%2即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int matrix[2000][2000]={0};
int t,n;

int lowbit(int x)
{
    return x&(-x);
}

int sum(int x,int y)
{
    int total=0;
    for (int i=x; i>0; i-=lowbit(i))
        for (int j=y; j>0; j-=lowbit(j))
            total+=matrix[i][j];
    return total;
}

void change(int x,int y)
{
    for (int i=x; i<=n; i+=lowbit(i))
        for (int j=y; j<=n; j+=lowbit(j))
            matrix[i][j]++;
}

int main()
{
    scanf("%d",&t);
    for (int T=1; T<=t; T++)
        {
            int m;
            scanf("%d%d",&n,&m);
            memset(matrix,0,sizeof(matrix));
            while (m>0)
                {
                    char command[10];
                    scanf("%s",&command);
                    if (command[0]=='C')
                        {
                            int x1,x2,y1,y2;
                            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                            change(x1,y1);
                            change(x2+1,y1);
                            change(x1,y2+1);
                            change(x2+1,y2+1);
                        }
                    if (command[0]=='Q')
                        {
                            int x,y;
                            scanf("%d%d",&x,&y);
                            int ans=sum(x,y) % 2;
                            printf("%d\n",ans);
                        }
                    m--;
                }
            printf("\n");
        }
    return 0;
}