POJMatrix（二维树状数组）

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 22058		Accepted: 8219

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
二维树状数组，跟一维的差不多，
这个道题的思路就是看看x1,y1往上加一，同时方块右边，下面和右下方的区域再加1，只要保证他们那边加个偶数就可以了。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int MAX =  + ;
 int c[MAX][MAX];
 int n;
 int lowbit(int k)
 {
     return k & (-k);
 }
 void update(int x,int y,int num)
 {
     for(int i = x; i < n; i += lowbit(i))
     {
         for(int j = y; j < n; j += lowbit(j))
             c[i][j] += num;
     }
 }
 int sum(int x,int y)
 {
     int s = ;
     for(int i = x; i > ; i -= lowbit(i))
     {
         for(int j = y; j > ; j -= lowbit(j))
             s += c[i][j];
     }
     return s;
 }
 int main()
 {
     int t,q;
     int num = ;
     scanf("%d", &t);
     while(t--)
     {
         if(num ++)
             printf("\n");
         scanf("%d%d", &n,&q);
         memset(c,,sizeof(c));
         char ch;
         int x1,y1,x2,y2;
         getchar();
         while(q--)
         {
             scanf("%c", &ch);
             if(ch == 'Q')
             {
                 scanf("%d%d", &x1,&y1);
                 getchar();
                 int m = sum(x2,y2) - sum(x1 - , y2) - sum(x2,y1 - ) + sum(x1-,y1-);
                 if(m %  == )
                     printf("0\n");
                 else
                     printf("1\n");
             }
             else if(ch == 'C')
             {
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 getchar();
                 update(x1,y1,);
             }
         }
     }
     return ;
 }