POJ2155(二维树状数组)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 17226		Accepted: 6461

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

 

题意：给出矩阵左上角和右下角坐标，矩阵里的元素 1变0 ，0 变1，然后给出询问，问某个点是多少

思路：二维树状数组

 //2017-10-25
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 const int N = ;

 int bt[N][N], n, q;

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

 void add(int x, int y, int v){
     while(x <= n){
         int j = y;
         while(j <= n){
             bt[x][j] += v;
             j += lowbit(j);
         }
         x += lowbit(x);
     }
 }

 int sum(int x, int y){
     int sm = ;
     while(x > ){
         int j = y;
         while(j > ){
             sm += bt[x][j];
             j -= lowbit(j);
         }
         x -= lowbit(x);
     }
     return sm;
 }

 int main()
 {
     int T;
     cin>>T;
     while(T--){
         scanf("%d%d", &n, &q);
         memset(bt, , sizeof(bt));
         char op;
         int x, y, x1, y1;
         while(q--){
             getchar();
             scanf("%c%d%d", &op, &x, &y);
             if(op == 'C'){
                 scanf("%d%d", &x1, &y1);
                 add(x, y, );
                 add(x, y1+, -);
                 add(x1+, y, -);
                 add(x1+, y1+, );
             }else{
                 printf("%d\n", sum(x, y)%);
             }
         }if(T)printf("\n");
     }

     return ;
 }