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.
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.
For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).
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
NO
YES
NO
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 #define PI acos(-1.0) 17 #define max(a,b) (a) > (b) ? (a) : (b) 18 #define min(a,b) (a) < (b) ? (a) : (b) 19 #define ll long long 20 #define eps 1e-10 21 #define MOD 1000000007 22 #define N 106 23 #define inf 1<<26 24 int n; 25 double x[N],y[N]; 26 double mp[N][N]; 27 double cal(int i,int j){ 28 return ((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])); 29 } 30 int main() 31 { 32 int t; 33 scanf("%d",&t); 34 while(t--){ 35 scanf("%d",&n); 36 for(int i=0;i<n;i++){ 37 scanf("%lf%lf",&x[i],&y[i]); 38 } 39 double minnn=inf; 40 for(int i=0;i<n;i++){ 41 for(int j=i+1;j<n;j++){ 42 mp[i][j]=cal(i,j); 43 if(minnn>mp[i][j]){ 44 minnn=mp[i][j]; 45 } 46 } 47 } 48 int ans=0; 49 for(int i=0;i<n;i++){ 50 for(int j=i+1;j<n;j++){ 51 if(mp[i][j]==minnn){ 52 ans++; 53 } 54 } 55 } 56 if(ans==n){ 57 printf("YES\n"); 58 }else{ 59 printf("NO\n"); 60 } 61 } 62 return 0; 63 }
hdu 5533 Dancing Stars on Me（数学，水）
原文：http://www.cnblogs.com/UniqueColor/p/4983298.html
