POJ

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 29525 Accepted: 10783

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

题意：给定一个矩阵 两个操作 一个是给出两个坐标 将坐标所形成的矩阵中的所有0变为1 1变为0 另一个是询问某个点是多少

思路：二维的树状数组

①

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <vector>#define max_ 1010#define inf 0x3f3f3f3f#define ll long longusing namespace std;int n,m;int c[max_][max_];int lb(int x){return (x&(-x));}void update(int x,int y,int v){for(int i=x;i<=n;i+=lb(i))for(int j=y;j<=n;j+=lb(j))c[i][j]+=v;}int getsum(int x,int y){int sum=0;for(int i=x;i>0;i-=lb(i))for(int j=y;j>0;j-=lb(j))sum+=c[i][j];return sum;}int main(int argc, char const *argv[]){int t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(c,0,sizeof(c));char c;int x1,x2,y1,y2;while(m--){scanf(" %c",&c);if(c=='C'){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);update(x2+1,y2+1,1);update(x1,y1,1);update(x1,y2+1,-1);update(x2+1,y1,-1);}else{scanf("%d%d",&x1,&y1);printf("%d\n",getsum(x1,y1)&1);}}printf("\n");}return 0;}

②

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <vector>#define max_ 1010#define inf 0x3f3f3f3f#define ll long longusing namespace std;int n,m;int c[max_][max_];int lb(int x){return (x&(-x));}void update(int x,int y){for(int i=x;i<=n;i+=lb(i))for(int j=y;j<=n;j+=lb(j))c[i][j]++;}int getsum(int x,int y){int sum=0;for(int i=x;i>0;i-=lb(i))for(int j=y;j>0;j-=lb(j))sum+=c[i][j];return sum;}int main(int argc, char const *argv[]){int t;scanf("%d",&t);while(t--){memset(c,0,sizeof(c));scanf("%d%d",&n,&m);char c;int x1,x2,y1,y2;while(m--){scanf(" %c",&c);if(c=='C'){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);x1++;x2++;y1++;y2++;update(x2,y2);update(x1-1,y2);update(x2,y1-1);update(x1-1,y1-1);}else if(c=='Q'){scanf("%d%d",&x1,&y1);printf("%d\n",getsum(x1,y1)&1);}}printf("\n");}return 0;}