2017 Multi-University Training Contest - Team 2

最新推荐文章于 2025-08-23 19:36:46 发布

weixin_34161029

最新推荐文章于 2025-08-23 19:36:46 发布

阅读量63

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/BobHuang/p/7246539.html

本文介绍了一个关于二维平面上寻找由给定点构成的不同正多边形数量的问题。通过输入一系列整数坐标点，利用算法计算并输出可能形成的正多边形数目。示例中特别提到正方形是最常见的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;
}