poj 2155 Matrix
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Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 25517 Accepted: 9450
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.
- 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).
- 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
【分析】
二维树状数组经典模板题
【代码】
//poj 2155 Matrix#include<iostream>#include<cstdio>#include<cstring>#define M(a) memset(a,0,sizeof a)#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;int n,t,T;int c[1005][1005];int lowbit(int x) {return x&(-x);}void add(int x,int y){ for(int i=x;i<=n;i+=lowbit(i)) for(int j=y;j<=n;j+=lowbit(j)) c[i][j]++;}int query(int x,int y){ int ans=0; for(int i=x;i;i-=lowbit(i)) for(int j=y;j;j-=lowbit(j)) ans+=c[i][j]; return ans%2;}int main(){ int i,j,k,x1,y1,x2,y2,x,y; scanf("%d",&T); while(T--) { M(c); scanf("%d%d",&n,&t); while(t--) { char tmp[2]; scanf("%s",tmp); if(tmp[0]=='C') { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); add(x1,y1); add(x1,y2+1); add(x2+1,y1); add(x2+1,y2+1); } else { scanf("%d%d",&x,&y); printf("%d\n",query(x,y)); } } printf("\n"); } return 0;
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