HDU 5533 Dancing Stars on Me(判断n个整数点是否能组成正n边形)——2015ACM/ICPC亚洲区长春站

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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

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

 

Sample Output

NOYESNO

 

/*********************************************************************/

题意：给你n个整点的坐标，问是否能组成正n边形。

解题思路：其实这道题，做过bestcoder的人应该多多少少有些印象吧，有一道类似的题，只不过那时候只问能不能组成正三角形、正方形、正五边形和正六边形，具体可查看链接

HDU 5365 Run ——BestCoder Round #50(div.1 div.2)

所以我们便可以得到格点(即整点，就是横纵坐标均为整数)只能构成正方形的结论

这么说来，我们只需要注意n=4才有可能是正n边形

也就是说只要我们判断4个点能不能组成正方形就可以了

#pragma comment(linker, "/STACK:1024000000,1024000000")#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>#define exp 1e-10using namespace std;const int N = 105;const int M = 2010;const int inf = 2147483647;const int mod = 2009;int x[N],y[N],L[6];int main(){     int  t,i,j,n,k;     scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0;i<n;i++)            scanf("%d%d",&x[i],&y[i]);        if(n!=4)        {            puts("NO");            continue;        }        for(k=i=0;i<n;i++)            for(j=0;j<i;j++,k++)                L[k]=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);        sort(L,L+6);        if(L[0]==L[1]&&L[2]==L[1]&&L[2]==L[3]&&L[4]==L[5]&&L[4]!=L[3])            puts("YES");        else            puts("NO");    }    return 0; }

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