POJ 2155 Matrix

二维树状数组。。。。

                         Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 15575		Accepted: 5854

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

Source

POJ Monthly,Lou Tiancheng

 #include <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 int tree[][],n,m,t;

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

 void update(int x,int y,int v)
 {
     int temp=y;
     while(x<=n)
     {
         y=temp;
         while(y<=n)
         {
             tree[x][y]+=v;
             y+=lowbit(y);
         }
         x+=lowbit(x);
     }
 }

 int getsum(int x,int y)
 {
     int sum=;
     int temp=y;
     while(x>)
     {
         y=temp;
         while(y>)
         {
             sum+=tree[x][y];
             y-=lowbit(y);
         }
         x-=lowbit(x);
     }
     return sum;
 }

 int main()
 {
     scanf("%d",&t);
     while(t--)
     {
         memset(tree,,sizeof(tree));
         scanf("%d%d",&n,&m);
         n++;char str[];
         while(m--)
         {
             scanf("%s",str);
             if(str[]=='C')
             {
                 int x1,x2,y1,y2;
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 x1++;y1++;x2++;y2++;
                 update(x1,y1,);
                 update(x1,y2+,-);
                 update(x2+,y1,-);
                 update(x2+,y2+,);
             }
             else if(str[]=='Q')
             {
                 int x,y;
                 scanf("%d%d",&x,&y);
                 x++;y++;
                 printf("%d\n",getsum(x,y)&);
             }
         }
         putchar();
     }
     return ;
 }