Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16797		Accepted: 6312

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]. 

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.

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.

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

1

0

0

1

POJ Monthly,Lou Tiancheng

 //4100K    438MS    C++    1052B    2014-04-01 21:00:26

 /*

     题意：

         给出一个n*n的 二维区间，有小矩形取反操作和对某元素求值操作，对每个操作

     做出处理。

     树状数组：

         这题和一般的树状数组不一样，一般的是对某个值更改，对某段求值，现在是刚好

     反过来，利用反向思维，讲原先更新的操作用来求值， 求值的操作用来更新即可。 

 */

 #include<stdio.h>

 #include<string.h>

 #define N 1005

 int c[N][N];

 int lowbit(int i)

 {

     return i&(-i);

 }

 int update(int x,int y)

 {

     int s=;

     for(int i=x;i<N;i+=lowbit(i))

         for(int j=y;j<N;j+=lowbit(j))

             s+=c[i][j];

     return s%;

 }

 void getsum(int x,int y)

 {

     for(int i=x;i>;i-=lowbit(i))

         for(int j=y;j>;j-=lowbit(j))

             c[i][j]^=;

 }

 int main(void)

 {

     int t,n,m;

     char op;

     int x1,x2,y1,y2;

     scanf("%d",&t);

     while(t--)

     {

         scanf("%d%d%*c",&n,&m);

         memset(c,,sizeof(c));

         while(m--){

             scanf("%c",&op);

             if(op=='Q'){

                 scanf("%d%d%*c",&x1,&y1);

                 printf("%d\n",update(x1,y1));

             }else{

                 scanf("%d%d%d%d%*c",&x1,&y1,&x2,&y2);

                 getsum(x1-,y1-);

                 getsum(x1-,y2);

                 getsum(x2,y1-);

                 getsum(x2,y2);

             }

         }

         printf("\n");

     }

     return ;

 }