hdu5563 Clarke and five-pointed star

思路：正五边形的判断。在正五边形中任意两点之间的距离只有两种情况，所以求出所有点之间的距离排序，前后比较纪录有多少个不同距离的边长。

// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
// #define DEBUG
#ifdef DEBUG
#define debug(...) printf( __VA_ARGS__ )
#else
#define debug(...)
#endif
#define MEM(x,y) memset(x, y,sizeof x)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<double,double> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
ii p[6];
const double eps = 1e-6;
double mul(double x){
	return x*x;
}
double GetDis(const ii& A,const ii& B){
	return sqrt(mul(A.first - B.first) + mul(A.second - B.second));
}
double dis[12];
int main()
{	
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int t;
	scanf("%d",&t);
	while(t--){
		int cnt = 1;
		for (int i = 1;i <= 5;++i)
			scanf("%lf%lf",&p[i].first,&p[i].second);
		for (int i = 1;i <= 5;++i){
			for (int j = i + 1;j <= 5;++j)
				dis[cnt++] = GetDis(p[i], p[j]);
		}
		sort(dis + 1,dis + cnt);
		int ans = 1;
		for (int i = 2;i < cnt;++i){
			if (fabs(dis[i] - dis[i-1]) > eps) ans++;
		}
		if (ans <= 2) puts("Yes");
		else puts("No");
	}
	return 0;
}