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[1111][1111];
int n,m;
char s[10];

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

int get(int x,int y)   
{
    int sum=0;
    for(int i=x; i>0; i-=lowbit(i)){
        for(int j=y; j>0; 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,0,sizeof(a));
        while(m--){
            scanf("%s",s);
            if(s[0]=='C'){  
                int x1,y1,x2,y2;  
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);  
                add(x2+1,y2+1);  
                add(x1,y1);  
                add(x1,y2+1);  
                add(x2+1,y1);  
            }  
            else{
                int x,y;
                scanf("%d%d",&x,&y);
                printf("%d
",get(x,y)%2);
            }
        }
        if(t)
            printf("
");
    }
    return 0;
}