Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a learner of geometric. 

When he did a research with polygons, he found he has to judge if the polygon is a five-pointed star at many times. There are 5 points on a plane, he wants to know if a five-pointed star existed with 5 points given.

 

Input

The first line contains an integer T(1≤T≤10),
the number of the test cases. 

For each test case, 5 lines follow. Each line contains 2 real numbers xi,yi(−109≤xi,yi≤109),
denoting the coordinate of this point.

 

Output

Two numbers are equal if and only if the difference between them is less than 10−4. 

For each test case, print Yes if
they can compose a five-pointed star. Otherwise, print No.
(If 5 points are the same, print Yes.
)

 

Sample Input

2

3.0000000 0.0000000

0.9270509 2.8531695

0.9270509 -2.8531695

-2.4270509 1.7633557

-2.4270509 -1.7633557

3.0000000 1.0000000

0.9270509 2.8531695

0.9270509 -2.8531695

-2.4270509 1.7633557

-2.4270509 -1.7633557

 

Sample Output

Yes

No

Hint






/*

  *题目大意：给你五个点的坐标、要求判断是否可以组成五角星

 *算法分析：注意在五点相同时候为YES，否则判断是否存在有两组五条相等的边， 存在则YES，否则NO

*/

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <cstdlib>

#include <cmath>

#include <algorithm>

using namespace std;

struct node{

	double x, y;

}a[5];

int panDuan(double a, double b) {

	if (fabs(a-b)<=1e-4)

		return 1;

	return 0;

}

double juLi(double x1, double y1, double x2, double y2) {

	return (double)(x1-x2)*(x1-x2) + (y1-y2)*(y1-y2);

}

int main() {

	int t;

	cin >> t;

	while (t --) {

		int flag = 0;

		memset(a, 0, sizeof(a));

		for (int i = 0; i<5; i++)

			cin >> a[i].x >> a[i].y;

		for (int i = 0; i<4; i++) {

			if (panDuan(a[i].x, a[i+1].x) == 0 || panDuan(a[i].y, a[i+1].y) == 0)

				flag = 1;

		}

		if (flag == 0)

			cout << "Yes" << endl;

		else {

			flag = 0;

			double ans1;

			double ans = juLi(a[0].x, a[0].y, a[1].x, a[1].y);

			for (int i = 0; i<5; i++) {

				for (int j = i+1; j<5; j++) {

					if (fabs(juLi(a[i].x, a[i].y, a[j].x, a[j].y) - ans) > 1e-4)

						ans1 = juLi(a[i].x, a[i].y, a[j].x, a[j].y);

				}

			}

			int flag1 = 0;

			//cout << ans << endl<< endl;

			for (int k = 0; k<5; k++) {

				for (int l = k+1; l<5; l++) {

					//cout << juLi(a[k].x, a[k].y, a[l].x, a[l].y) << endl << endl;

					if (panDuan(juLi(a[k].x, a[k].y, a[l].x, a[l].y), ans) == 1)

						flag ++ ;

					if (panDuan(juLi(a[k].x, a[k].y, a[l].x, a[l].y), ans1) == 1)

						flag1 ++ ;

				}

			}

			//cout << flag << endl;

			if (flag == 5 && flag1 == 5)

				cout << "Yes" << endl;

			else

				cout << "No" << endl;

		}

	}

	return 0;

}