Matrix (二维树状数组)

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

题解:这个PDF讲的挺好->   浅谈信息学竞赛中的“0 ” 和 “1” ——二进制思想在信息学竞赛中的应用

代码:

 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <algorithm>
 #include <iostream>
 #include <ctype.h>
 #include <iomanip>
 #include <queue>
 #include <stdlib.h>
 using namespace std;

 int a[][];
 int n,m;
 char s[];

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

 int get(int x,int y)
 {
     int sum=;
     for(int i=x; i>; i-=lowbit(i)){
         for(int j=y; j>; j-=lowbit(j)){
             sum+=a[i][j];
         }
     }
     return sum;
 } 

 void add(int x,int y)
 {
     for(int i=x; i<=n; i+=lowbit(i)){
         for(int j=y; j<=n; j+=lowbit(j)){
             a[i][j]++;
         }
     }
 }

 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&m);
         memset(a,,sizeof(a));
         while(m--){
             scanf("%s",s);
             if(s[]=='C'){
                 int x1,y1,x2,y2;
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 add(x2+,y2+);
                 add(x1,y1);
                 add(x1,y2+);
                 add(x2+,y1);
             }
             else{
                 int x,y;
                 scanf("%d%d",&x,&y);
                 printf("%d\n",get(x,y)%);
             }
         }
         if(t)
             printf("\n");
     }
     return ;
 }