标签：Me HDU polygon form Dancing sky regular stars include

The sky was brushed clean by the wind and the stars were cold in a black sky. What a wonderful night. You observed that, sometimes the stars can form a regular polygon in the sky if we connect them properly. You want to record these moments by your smart camera. Of course, you cannot stay awake all night for capturing. So you decide to write a program running on the smart camera to check whether the stars can form a regular polygon and capture these moments automatically.

Formally, a regular polygon is a convex polygon whose angles are all equal and all its sides have the same length. The area of a regular polygon must be nonzero. We say the stars can form a regular polygon if they are exactly the vertices of some regular polygon. To simplify the problem, we project the sky to a two-dimensional plane here, and you just need to check whether the stars can form a regular polygon in this plane.

InputThe first line contains a integer TT indicating the total number of test cases. Each test case begins with an integer nn, denoting the number of stars in the sky. Following nn lines, each contains 22 integers xi,yixi,yi, describe the coordinates of nn stars.

1≤T≤3001≤T≤300
3≤n≤1003≤n≤100
−10000≤xi,yi≤10000−10000≤xi,yi≤10000
All coordinates are distinct.OutputFor each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).Sample Input

3
3
0 0
1 1
1 0
4
0 0
0 1
1 0
1 1
5
0 0
0 1
0 2
2 2
2 0

Sample Output

NO
YES
NO
思路：多边形中，n个点对任意其他点的距离，总共有n/2种不同的长度，利用这个规则，可直接求。

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
using namespace std;

int t, n, x[100+8], y[100+8];
map<int, int>len;

int main()
{
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        len.clear();//一定要记得清空，不然size会改变，插入的时候也会继续往后插
        int ii = 0;
        for(int i = 0; i<n; i++)
            scanf("%d%d", &x[i], &y[i]);
        for(int i = 0; i<n; i++)
            for(int j = i+1; j<n; j++)
                len[(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])] = 1;
        if(len.size() == n/2)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

 

标签：Me,HDU,polygon,form,Dancing,sky,regular,stars,include
来源： https://www.cnblogs.com/RootVount/p/10908440.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。