HDU-5533 Dancing Stars on Me

Posted 2020-08-09 Kiven#5197

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1175    Accepted Submission(s): 643

 

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 nlines, 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：

求所给点能否组成正n变形。

 

因为给定的点都为整数，故只有可能组成正四边形即正方形。

我们只要判断有多少边==最小边，若结果为n，则YES。

 

附AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int INF=1<<30;
 5 const int MAXN=110; 
 6 
 7 double x[MAXN],y[MAXN];
 8 
 9 double Len(int i,int j){
10     return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
11 }
12 
13 int main(){
14     int t;
15     cin>>t;
16     while(t--){
17         int n;
18         cin>>n;
19         for(int i=1;i<=n;i++){
20             cin>>x[i]>>y[i];
21         }
22         double MIN=INF;
23         for(int i=1;i<=n;i++){
24             for(int j=i+1;j<=n;j++){
25                 MIN=min(Len(i,j),MIN);
26             }
27         }
28         int ans=0;
29         for(int i=1;i<=n;i++){
30             for(int j=i+1;j<=n;j++){
31                 if(Len(i,j)==MIN)
32                 ans++;
33             }
34         }
35         if(ans==n)
36         cout<<"YES"<<endl;
37         else
38         cout<<"NO"<<endl;
39     }
40     return 0;
41 }