2017 Multi-University Training Contest

Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 177    Accepted Submission(s): 74

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

4 0 0 0 1 1 0 1 1 6 0 0 0 1 1 0 1 1 2 0 2 1

 

Sample Output

1 2

 

Source

2017 Multi-University Training Contest - Team 2

 

最弱的一题，不过这个正多边形必然是正方形有点骚，不好想啊

#include <bits/stdc++.h>
using namespace std;
int main() {
    int x[505],y[505];
    int n;
    while(cin>>n) {
        set<pair<int,int> >S;
        for(int i=0; i<n; i++) {
            cin>>x[i]>>y[i];
            S.insert(make_pair(x[i],y[i]));
        }
        int cnt=0;
        for(int i=0; i<n; i++)
            for(int j=0; j<i; j++) {
                int mx=x[i]+x[j],dx=max(x[i],x[j])-min(x[i],x[j]);
                int my=y[i]+y[j],dy=max(y[i],y[j])-min(y[i],y[j]);
                if ((mx+dy)&1||(my+dx)&1) continue;
                int sg=((x[i]-x[j])*(y[i]-y[j])<0)?1:-1;
                if (S.count(make_pair((mx+dy)/2,(my+sg*dx)/2))&&S.count(make_pair((mx-dy)/2,(my-sg*dx)/2)))
                    cnt++;

            }
        cout<<cnt/2<<endl;
    }
    return 0;
}