poj2155Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 20160 Accepted: 7531

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

Source

POJ Monthly,Lou Tiancheng

题意是给你个矩阵里面开始全是0，然后给你两种指令：1：‘C x1,y1,x2,y2’就是将左上角为x1,y1,右下角为x2,y2,的这个矩阵内的数字全部翻转，0变1,1变0,；2:'Q x1 y1',输出a[x1][y1]的值。

二维树状数组：对于一个矩阵,迅速修改其一个点的值,迅速求其左上角连续子矩阵的和

#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;int bit[1010][1010],n;void update(int r,int c,int val){for(int i=r;i<=n;i+=i&-i)for(int j=c;j<=n;j+=j&-j)bit[i][j]+=val;}int psum(int r,int c){int ans=0;for(int i=r;i>0;i-=i&-i)for(int j=c;j>0;j-=j&-j)ans+=bit[i][j];return ans;}int main(){int T;scanf("%d",&T);while(T--){int m;scanf("%d%d",&n,&m);memset(bit,0,sizeof(bit));while(m--){char s[100];scanf("%s",s);if(s[0]=='C'){int r1,c1,r2,c2;scanf("%d%d%d%d",&r1,&c1,&r2,&c2);update(r1,c1,1);update(r1,c2+1,-1);update(r2+1,c1,-1);update(r2+1,c2+1,1);}else{int r,c;scanf("%d%d",&r,&c);printf("%d\n",psum(r,c)%2);}}puts("");}}