HDU-5533 Dancing stars on me

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 integerT 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

330 01 11 040 00 11 01 150 00 10 22 22 0

Sample Output

NOYESNO
题目大意：判断给出的坐标可不可以构成正m边形

首先要知道，因为坐标是整数,只有当 m == 4 的时候才有可能是正方形，别的都不可能
所以只要判断是不是正方形就 ok 了， 可是给出的坐标不一定是按照顺序给出的，所以要判断一下，让他按照顺序给出

#include <cstdio>#define MAXN 1100#define INF 0x3f3f3f3fdouble x[MAXN],y[MAXN];double map[MAXN][MAXN];double fun(int i,int j){return (x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);}int main(){int T;int N;int i,j;scanf("%d",&T);while(T--){scanf("%d",&N);for(i=1;i<=N;++i){scanf("%lf%lf",x+i,y+i);}double min=INF;for(i=1;i<=N;++i){for(j=i+1;j<=N;++j){map[i][j]=fun(i,j);if(min>map[i][j]){min=map[i][j];}}}int cnt=0;for(i=1;i<=N;++i)for(j=i+1;j<=N;++j)if(min==map[i][j])cnt++;if(cnt==N) printf("YES\n");else printf("NO\n");}return 0;}