hdu 5533 计算几何 判断是否为正方形

原题目链接：

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1797    Accepted Submission(s): 1045

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

最后题目意思就是给你n个点，判断是否能组成正n边形。实际上由于正三角形正五边形，正六边形不可能都由整数点（即横坐标和纵坐标都是整数）构成，故我们只用判断能组成正方形即可。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
struct point
{
    int  x,y;
} pp[110];
int main()
{
    int  t,i,j,n,k,cnt;
    int  s[6];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        cnt=0;
        for(i=0; i<n; i++)
            scanf("%d%d",&pp[i].x,&pp[i].y);
        if(n!=4)
        {
            puts("NO");
            continue;
        }
        for(i=0; i<n; i++)
        {
            for(j=0; j<i; j++)
            {
                s[cnt++]=(pp[i].x-pp[j].x)*(pp[i].x-pp[j].x)+(pp[i].y-pp[j].y)*(pp[i].y-pp[j].y);
            }
        }

 sort(s,s+6);
        if(s[0]==s[1]&&s[2]==s[1]&&s[2]==s[3]&&s[4]==s[5]&&s[4]!=s[3])
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}