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http://poj.org/problem?id=2155
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.
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
这是一题简单的二维树状数组。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int maxn=1005; int va[maxn][maxn]; int n,m; int lowbit(int x) { return x&(-x); } void update(int x,int y) { for(int i=x;i<=n;i+=lowbit(i)) for(int j=y;j<=n;j+=lowbit(j)) va[i][j] +=1; } int Sum(int x,int y) { int ans=0; for(int i=x;i>0;i-=lowbit(i)) for(int j=y;j>0;j-=lowbit(j)) ans+=va[i][j]; return ans; } int main() { int N; int x1,y1,x2,y2; char op[10]; scanf("%d",&N); while(N--) { memset(va,0,sizeof(va)); scanf("%d %d",&n,&m); while(m--) { scanf("%S",op); if(op[0]=='C') { scanf("%d %d %d %d",&x1,&y1,&x2,&y2); update(x1,y1); update(x1,y2+1); update(x2+1,y1); update(x2+1,y2+1); } else { scanf("%d %d",&x1,&y1); printf("%d\n",Sum(x1,y1)%2); } } printf("\n"); } return 0; }
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原文地址:http://blog.csdn.net/hellohelloc/article/details/47699699