HDU_5533_Dancing Stars on Me

Posted 2020-10-09 edward108

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2002    Accepted Submission(s): 1168

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

 

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

• 给出n个二维坐标点判断能够构成正多边形
• 考虑正多边形顶点在同一个圆上
• 任意挑3个点找外心，得到一组圆心和半径
• 先对于每个点检测和圆心的距离
• 然后如果距离都相等那就判断每两个相邻点的距离也应该相等
• n的范围不大n^2复杂度判相邻即可

 

 

 1 #include <iostream>
 2 #include <string>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <cmath>
 8 #include <vector>
 9 #include <queue>
10 #include <stack>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long           LL ;
15 typedef unsigned long long ULL ;
16 const int    maxn = 1e2 + 10   ;
17 const int    inf  = 0x3f3f3f3f ;
18 const int    npos = -1         ;
19 const int    mod  = 1e9 + 7    ;
20 const int    mxx  = 100 + 5    ;
21 const double eps  = 1e-6       ;
22 const double PI   = acos(-1.0) ;
23 
24 struct node{
25     double x, y;
26     node(double a=0LL, double b=0LL){
27         x=a; y=b;
28     }
29 };
30 node vex[maxn], center;
31 node CircumCenter(node a, node b, node c){
32     double a1=b.x-a.x, b1=b.y-a.y, c1=(a1*a1+b1*b1)/2;
33     double a2=c.x-a.x, b2=c.y-a.y, c2=(a2*a2+b2*b2)/2;
34     double d=a1*b2-a2*b1;
35     return node(a.x+(c1*b2-c2*b1)/d,a.y+(a1*c2-a2*c1)/d);
36 }
37 double dis2(node a, node b){
38     return pow(a.x-b.x,2)+pow(a.y-b.y,2);
39 }
40 bool oneLine(node a, node b, node c){
41     return ((b.y-a.y)/(b.x-a.x))==((c.y-a.y)/(c.x-a.x));
42 }
43 int T, n, ans;
44 double R, D[maxn][maxn];
45 int main(){
46     // freopen("in.txt","r",stdin);
47     // freopen("out.txt","w",stdout);
48     while(~scanf("%d",&T)){
49         while(T--){
50             ans=1;
51             scanf("%d",&n);
52             for(int i=1;i<=n;i++)
53                 scanf("%lf %lf",&vex[i].x,&vex[i].y);
54             
55             if(oneLine(vex[1],vex[2],vex[3]))
56                 ans=0;
57             
58             if(ans){
59                 center=CircumCenter(vex[1],vex[2],vex[3]);
60                 R=dis2(center,vex[1]);
61                 for(int i=1;i<=n;i++)
62                     if(dis2(vex[i],center)!=R){
63                         ans=0; break;
64                     }
65             }
66 
67             if(ans){
68                 for(int i=1;i<=n;i++){
69                     D[i][i]=0.0;
70                     for(int j=i+1;j<=n;j++)
71                         D[i][j]=D[j][i]=dis2(vex[i],vex[j]);
72                     sort(D[i]+1,D[i]+1+n);
73                 }
74                 for(int i=2;i<=n;i++)
75                     if(!(D[i][2]==D[i][3] && D[i][2]==D[i-1][2])){
76                         ans=0; break;
77                     }
78             }
79             puts(ans?"YES":"NO");
80         }
81     }
82     return 0;
83 }