判断四个点是否能构成正方形

代码：

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

struct point
{
	double x, y;
} a[4];

bool Compare(point a, point b)
{
	if (a.x != b.x)
		return a.x < b.x; //如果，横坐标不相等，所有点按横坐标升序排列
	return a.y < b.y;//如果横坐标相等，所有点按纵坐标升序排列
}

double dis(point a, point b)//计算两点之间的距离
{
	return sqrt((a.x - b.x)*(a.x - b.x)) + (a.y - b.y)*(a.y - b.y));
}

bool IsRightAngle(point a, point b, point c)//判断是否为直角
{
	double x;
	x = (a.x - b.x)* (a.x - c.x) + (a.y - b.y)*(a.y - c.y);
	if (x < 0.00001)
		return 1;
	else
		return 0;
}

int main()
{
	int n, k;
	double s1, s2, s3, s4;

	scanf("%d",&n);

	while (n--)
	{
		for (int i = 0; i < 4; i++)
			cin >> a[i].x >> a[i].y;

		//确定点，排序，给点确定标号
		sort(a, a + 4, Compare);

		//确定边
		s1 = dis(a[0], a[2]);
		s2 = dis(a[0], a[1]);
		s3 = dis(a[3], a[1]);
		s4 = dis(a[2], a[3]);

		//分析是否为正方形
		if (s1 == s2&&s3 == s4&&s1 != 0 && IsRightAngle(a[0], a[1], a[2]))//三个条件同时满足（1：四条边相等，2：边不为0，3：有一个直角）
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
		cout << endl;
		
	}

	return 0;
}