HDU 5533 Dancing Stars on Me （2015ACM/ICPC亚洲区长春计算几何）

本文主要是介绍HDU 5533 Dancing Stars on Me （2015ACM/ICPC亚洲区长春计算几何），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

【题目链接】：click here~~

【题目描述】:

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 141 Accepted Submission(s): 96

Problem Description

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.

Input

The first line contains a integer

T indicating the total number of test cases. Each test case begins with an integer

n , denoting the number of stars in the sky. Following

n lines, each contains

2 integers

xi,yi , describe the coordinates of

n stars.

1≤T≤300

3≤n≤100

−10000≤xi,yi≤10000
All coordinates are distinct.

Output

For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).

Sample Input

Sample Output

  NO
YES
NO

【题意】：给出n个点的坐标，问能否组成正多边形。

【思路】判断边长是否相等，对角线相等。。

代码：

/*
* Problem:HDU 5533
* Running time: 15MS
* Complier: G++
* Author: javaherongwei
* Create Time: 21:20 2015/11/02 星期一
*/
#include <iostream>
#include <stdio.h>
#include <bits/stdc++.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=233;
int dir4[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
int mat[maxn][maxn];
typedef pair<int,int>pii;
pii p1[maxn],p2[maxn];
int dis(pii a,pii b)
{int x=a.first-b.first;int y=a.second-b.second;return x*x+y*y;
}
void solve()
{int ok=0;int ck[]= {0,1,2,3};do{for(int i=0; i<4; ++i){p2[i]=p1[ck[i]];}bool okk=1;for(int i=1; i<4; ++i){if(dis(p2[i],p2[(i+1)%4])!=dis(p2[0],p2[1])){okk=0;break;}}for(int i=0; i<2; ++i){if(dis(p2[i],p2[(i+2)])!=2*dis(p2[0],p2[1])){okk=0;break;}}if(!okk) continue;if(okk){ok=1;break;}}while(next_permutation(ck,ck+4));if(ok) puts("YES");else puts("NO");
}
int main()
{//freopen("1.txt","r",stdin);int t ;scanf("%d",&t);while(t--){int n;scanf("%d",&n);for(int i=0; i<n; ++i){scanf("%d%d",&p1[i].first,&p1[i].second);}solve();}return 0;
}

这篇关于HDU 5533 Dancing Stars on Me （2015ACM/ICPC亚洲区长春计算几何）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！