hdu 5533 Dancing Stars on Me

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5533

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 601    Accepted Submission(s):
320

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

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

 

题意：给你n个坐标让你判断是不是正n边形；

题解：从第一个点a开始，找这个点到其余所有点中距离最短的点b，然后将点a标记，再以b为起点找其余没有标记过的点中距离他最近的c，再将b点标记，以此类推，（注意将所有的最短边存下来）最后如果所有的最短边都相等则是正多边形

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
#include<math.h>
#define MAX 10010
#define INF 0x3f3f3f
#define DD double
using namespace std;
DD f(DD x1,DD y1,DD x2,DD y2)
{
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
	int t,n,m,j,i,k;
	DD x[MAX],y[MAX];
	DD s[MAX];
	int vis[MAX];
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			scanf("%lf%lf",&x[i],&y[i]);
		int k=0;
		memset(vis,0,sizeof(vis));
		DD Min;
		int next=1;
		int ans=1;
		for(i=1;i<=n;i++)
		{
			Min=INF;
			for(j=1;j<=n;j++)
			{
				if(next==j) continue;
				//如果自己到自己就跳过
				else if(!vis[j])
				{
					if(Min>f(x[next],y[next],x[j],y[j]))
					{
						Min=f(x[next],y[next],x[j],y[j]);
						//找距离next点最近的点
						ans=j;
					}
				}
			}
			next=ans;   //找到下一个点
			vis[next]=1;
			s[k++]=Min;
		}
		int flag=1;
		for(i=0;i<k-1;i++)
		{
			if(s[i]!=s[i+1])
			{
				flag=0;
				break;
			}
		}
		if(s[0]!=s[k-1])
		    flag=0;
		if(flag)
		printf("YES\n");
		else
		printf("NO\n");
	}
	return 0;
}