POJ-2155 Matrix (二维树状数组 入门题)

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 29734 Accepted: 10865

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

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

#include <iostream>  #include <cstdio>  #include <cstdlib>  #include <cstring>  #include <string>  #include <algorithm>using namespace std;int c[1002][1002];inline int lowbit(int x){return x & (-x);}void add(int x, int y, int n, int v){while(x <= n){int cur = y;while(cur <= n){c[x][cur] += v;cur += lowbit(cur);}x += lowbit(x);}}int query(int x, int y){int ans = 0;while(x){int cur = y;while(cur){ans += c[x][cur];cur -= lowbit(cur);}x -= lowbit(x);}return ans;}int main(){int T;scanf("%d", &T);while(T--){memset(c, 0, sizeof(c));int n, q, x1, x2, y1, y2;scanf("%d %d", &n, &q);char s;for(int i = 1; i <= q; ++i){getchar();s = getchar();if(s == 'C'){scanf("%d %d %d %d", &x1, &y1, &x2, &y2);add(x1, y1, n, 1);add(x1, y2 + 1, n, 1);add(x2 + 1, y1, n, 1);add(x2 + 1, y2 + 1, n, 1);}else{scanf("%d %d", &x1, &y1);printf("%d\n", query(x1, y1) % 2);}}if(T)printf("\n");}}/*题意：1000*1000的网格，5e4次操作，每次操作要么改变一个区块的数值，0改成1,1改成0；要么询问某点上得数字。思路：暴力的话是1000*1000*50000，突破天际。从复杂度来源上来看，可以在区块修改上做点文章。每个点上的数字只有0和1两种情况，每次修改是对整个区块做一个翻转，那么我们似乎只要做一个区块的翻转标记就好了，而不需要整体修改。按照题中的变量，假设原点在左上角，我们在(x1,y1)处做一个翻转标记，这个标记代表左上角为(x1,y1)，右下角无限延生的一个区块都做了翻转。那么我们利用这个假设对一个有限区块做翻转标记只需要标记四个点就可以了？显然是的，那么怎么询问某个点(x, y)上当前的值呢？我们只需要统计(0, 0)到(x, y)中有多少个翻转标记即可，再结合初始值算一下。这个统计显然可以用二维树状数组来做统计。*/