HDU5533 Dancing Stars on Me

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1271 Accepted Submission(s): 702

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）

Recommend
hujie

题意：给出n各点的坐标，判断其能否构成正n边形。
解题思路：点数最大只有100，暴力枚举两点之间的距离（简化起见，没有开方），若其中最小距离的个数等于n，则YES，否则NO。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;struct point{    int x,y;}p[105];int dis(point p1,point p2){    return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);}int main(){    int T;    scanf("%d",&T);    int n;    while(T--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d%d",&(p[i].x),&(p[i].y));        }        int minlen=0x3f3f3f3f;        int mincnt=0;        for(int i=1;i<n;i++)        {            for(int j=i+1;j<=n;j++)            {                if(minlen>dis(p[i],p[j]))                {                    minlen=dis(p[i],p[j]);                    mincnt=1;                }                else if(minlen==dis(p[i],p[j]))                {                    mincnt++;                }            }        }        if(mincnt==n)            printf("YES\n");        else            printf("NO\n");    }    return 0;}