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【题目链接】：click here~~
【题目描述】:
Dancing Stars on Me
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 141 Accepted Submission(s): 96
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.
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 x![]()
i![]()
,y![]()
i![]()
![]()
, describe the coordinates of n![]()
stars.
1≤T≤300![]()
3≤n≤100![]()
−10000≤x![]()
i![]()
,y![]()
i![]()
≤10000![]()
All coordinates are distinct.
1≤T≤300
3≤n≤100
−10000≤x
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
【题意】：给出n个点的坐标，问能否组成正多边形。
【思路】判断边长是否相等，对角线相等。。
代码：
/*
* Problem:HDU 5533
* Running time: 15MS
* Complier: G++
* Author: javaherongwei
* Create Time: 21:20 2015/11/02 星期一
*/
#include <iostream>
#include <stdio.h>
#include <bits/stdc++.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn=233;
int dir4[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};
int mat[maxn][maxn];
typedef pair<int,int>pii;
pii p1[maxn],p2[maxn];
int dis(pii a,pii b)
{int x=a.first-b.first;int y=a.second-b.second;return x*x+y*y;
}
void solve()
{int ok=0;int ck[]= {0,1,2,3};do{for(int i=0; i<4; ++i){p2[i]=p1[ck[i]];}bool okk=1;for(int i=1; i<4; ++i){if(dis(p2[i],p2[(i+1)%4])!=dis(p2[0],p2[1])){okk=0;break;}}for(int i=0; i<2; ++i){if(dis(p2[i],p2[(i+2)])!=2*dis(p2[0],p2[1])){okk=0;break;}}if(!okk) continue;if(okk){ok=1;break;}}while(next_permutation(ck,ck+4));if(ok) puts("YES");else puts("NO");
}
int main()
{//freopen("1.txt","r",stdin);int t ;scanf("%d",&t);while(t--){int n;scanf("%d",&n);for(int i=0; i<n; ++i){scanf("%d%d",&p1[i].first,&p1[i].second);}solve();}return 0;
}
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