hdu4720 三角形的最小圆覆盖

求一个三角形的最小圆覆盖

两种情况：

1. 是锐角及直角三角形， 那么这个圆就是外接圆

2. 是钝角三角形， 那么这个圆的直径是这个最长边的中点

double getDis(Point a, Point b)
{
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

///////////////////////////////////////////  
//求三角形外接圆圆心坐标 及半径
///////////////////////////////////////////  
void  circle_center(Point  ¢er,Point pt[3],double  &radiu)    //参数是圆心和半径的引用
{  
	double  x1,x2,x3,y1,y2,y3;  
	double  x  =  0;  
	double  y  =  0;  

	x1  = pt[0].x;  
	x2  = pt[1].x;  
	x3  = pt[2].x;  
	y1  = pt[0].y;  
	y2  = pt[1].y;  
	y3  = pt[2].y;  

	x=((y2-y1)*(y3*y3-y1*y1+x3*x3-x1*x1)-(y3-y1)*(y2*y2-y1*y1+x2*x2-x1*x1))/(2*(x3-x1)*(y2-y1)-2*((x2-x1)*(y3-y1)));  
	y=((x2-x1)*(x3*x3-x1*x1+y3*y3-y1*y1)-(x3-x1)*(x2*x2-x1*x1+y2*y2-y1*y1))/(2*(y3-y1)*(x2-x1)-2*((y2-y1)*(x3-x1)));  

	center.x  =  x  ;  
	center.y  =  y  ;  
	radiu  =  sqrt((pt[0].x  -  x)*(pt[0].x  -  x)  +  (pt[0].y  -  y)*(pt[0].y  -  y)); 
}

bool judgeDunJiao(Point pt[3], Point &center, double &r)   //判断是否是钝角， 参数是引用
{
	double c = getDis(pt[0], pt[1]);
	double b = getDis(pt[0], pt[2]);
	double a = getDis(pt[1], pt[2]);
	if(a * a > b * b + c * c)
	{
		center.x = (pt[1].x + pt[2].x) / 2.0;
		center.y = (pt[1].y + pt[2].y) / 2.0;
		r = a / 2.0;
		return true;
	}
	else if(b * b > a * a + c * c)
	{
		center.x = (pt[0].x + pt[2].x) / 2.0;
		center.y = (pt[0].y + pt[2].y) / 2.0;
		r = a / 2.0;
		return true;
	}
	else if(c * c > a * a + b * b)
	{
		center.x = (pt[1].x + pt[0].x) / 2.0;
		center.y = (pt[1].y + pt[0].y) / 2.0;
		r = a / 2.0;
		return true;
	}
	else
		return false;

}

参考题目：hdu4720

ac代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>

#define print(a) cout << #a << " : " << a << endl;
using namespace std;

const int M = 1003;
const double e = 1e-6;

struct Point
{
	double x, y;
};

Point wi[4];

double getDis(Point a, Point b)
{
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
///////////////////////////////////////////  
//求三角形外接圆圆心坐标 及半径
///////////////////////////////////////////  
void  circle_center(Point  ¢er,Point pt[3],double  &radiu)    //参数是圆心和半径的引用
{  
	double  x1,x2,x3,y1,y2,y3;  
	double  x  =  0;  
	double  y  =  0;  

	x1  = pt[0].x;  
	x2  = pt[1].x;  
	x3  = pt[2].x;  
	y1  = pt[0].y;  
	y2  = pt[1].y;  
	y3  = pt[2].y;  

	x=((y2-y1)*(y3*y3-y1*y1+x3*x3-x1*x1)-(y3-y1)*(y2*y2-y1*y1+x2*x2-x1*x1))/(2*(x3-x1)*(y2-y1)-2*((x2-x1)*(y3-y1)));  
	y=((x2-x1)*(x3*x3-x1*x1+y3*y3-y1*y1)-(x3-x1)*(x2*x2-x1*x1+y2*y2-y1*y1))/(2*(y3-y1)*(x2-x1)-2*((y2-y1)*(x3-x1)));  

	center.x  =  x  ;  
	center.y  =  y  ;  
	radiu  =  sqrt((pt[0].x  -  x)*(pt[0].x  -  x)  +  (pt[0].y  -  y)*(pt[0].y  -  y)); 
}

bool judgeDunJiao(Point pt[3], Point ¢er, double &r)
{
	double c = getDis(pt[0], pt[1]);
	double b = getDis(pt[0], pt[2]);
	double a = getDis(pt[1], pt[2]);
	if(a * a > b * b + c * c)
	{
		center.x = (pt[1].x + pt[2].x) / 2.0;
		center.y = (pt[1].y + pt[2].y) / 2.0;
		r = a / 2.0;
		return true;
	}
	else if(b * b > a * a + c * c)
	{
		center.x = (pt[0].x + pt[2].x) / 2.0;
		center.y = (pt[0].y + pt[2].y) / 2.0;
		r = a / 2.0;
		return true;
	}
	else if(c * c > a * a + b * b)
	{
		center.x = (pt[1].x + pt[0].x) / 2.0;
		center.y = (pt[1].y + pt[0].y) / 2.0;
		r = a / 2.0;
		return true;
	}
	else
		return false;

}

void solve()
{	
	double r;	
	Point center;
	int dun = judgeDunJiao(wi, center, r);
	if(dun)
	{
		if(getDis(wi[3], center) - r > 0)
			cout << "Safe" << endl;
		else
			cout << "Danger" << endl;
		return;
	}

	
	circle_center(center, wi, r);
	double dis = getDis(center, wi[3]);
	if(dis - r > 0)
		cout << "Safe" << endl;
	else
		cout << "Danger" << endl;
}
int main()
{
	int ncase;
	int tmp = 0;
	cin >> ncase;
	while(ncase--)
	{
		
		for(int i = 0; i < 4; i++)
			cin >> wi[i].x >> wi[i].y;
		printf("Case #%d: ", ++tmp);
		solve();
	}
	return 0;
};