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 = 1100;

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 = 0;
    while(x > 0){
        int j = y;
        while(j > 0){
            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, 0, 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, 1);
                add(x, y1+1, -1);
                add(x1+1, y, -1);
                add(x1+1, y1+1, 1);
            }else{
                printf("%d
", sum(x, y)%2);
            }
        }if(T)printf("
");
    }

    return 0;
}