2017多校训练Contest2: 1011 Regular polygon hdu6055

Problem Description

On a two-dimensional plane, give you n integer points. Your task is to figure out how many different regular polygon these points can make.

 

Input

The input file consists of several test cases. Each case the first line is a numbers N (N <= 500). The next N lines ,each line contain two number Xi and Yi(-100 <= xi,yi <= 100), means the points’ position.(the data assures no two points share the same position.)

 

Output

For each case, output a number means how many different regular polygon these points can make.

 

Sample Input

40 00 11 01 160 00 11 01 12 02 1

 

Sample Output

12

数正方形个数

直接枚举两个点后看另外两个点是否存在即可。最后把重复统计的次数除掉

#include<queue>#include<vector>#include<cstdio>#include<string>#include<cstring>#include<cassert>#include<iostream>#include<algorithm>using namespace std;int mp[801][801];int x[601],y[601];int main(){int n;while(scanf("%d",&n)!=EOF){memset(mp,0,sizeof(mp));int i,j;for(i=1;i<=n;i++){scanf("%d%d",&x[i],&y[i]);x[i]+=400;y[i]+=400;mp[x[i]][y[i]]=1;}int ans=0;for(i=1;i<=n;i++){for(j=i+1;j<=n;j++){//if(i==j)//continue;int dx=x[j]-x[i],dy=y[j]-y[i];//if(v[dx][dy]||v[dy][dx])//continue;//v[dx][dy]=true;//v[dy][dx]=true;if(mp[x[j]+dy][y[j]-dx]&&mp[x[i]+dy][y[i]-dx])ans++;if(mp[x[j]-dy][y[j]+dx]&&mp[x[i]-dy][y[i]+dx])ans++;}}printf("%d\n",ans/4);}return 0;}