poj2155一个二维树状数组

                                                                                                               Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 15125		Accepted: 5683

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

一个二维树状数组，好像题目说错了，其实是左下角和右上角的。

 #include <iostream>
 #include <algorithm>
 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 using namespace std;
 long long int a[][];
 int m;
 int lowbit(int x)
 {
     return x&(-x);
 }
 void update(int x,int y,int x1,int y1)
 {
     int i,j;
     for(i=x1;i>;i-=lowbit(i))
     for(j=y1;j>;j-=lowbit(j))
     a[i][j]++;

     for(i=x-;i>;i-=lowbit(i))
     for(j=y-;j>;j-=lowbit(j))
     a[i][j]++;

     for(i=x1;i>;i-=lowbit(i))
     for(j=y-;j>;j-=lowbit(j))
     a[i][j]++;

     for(i=x-;i>;i-=lowbit(i))
     for(j=y1;j>;j-=lowbit(j))
     a[i][j]++;
 }
 int fun(int x,int y)
 {
     int i,j;
     int sum=;
     for(i=x;i<=m;i+=lowbit(i))
     for(j=y;j<=m;j+=lowbit(j))
     sum+=a[i][j];
     return sum%;
 }
 int main()
 {
     //freopen("int.txt","r",stdin);
     int n;
     int i,j;
     scanf("%d",&n);
     for(i=;i<n;i++)
     {
         memset(a,,sizeof(a));
         int k;
         scanf("%d%d",&m,&k);
         char b;
         for(j=;j<k;j++)
         {
             cin>>b;
             int x1,x2,y1,y2;
             if(b=='Q')
             {
                 scanf("%d%d",&x1,&y1);
                 printf("%d\n",fun(x1,y1));
             }
             else
             {
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 update(x1,y1,x2,y2);
             }
         }
         if(i!=n-)
         printf("\n");
     }
 }