poj1673

所謂Exocenter就是垂心。不難證明。

 
#include <iostream>   
#include <math.h>   
#include <stdio.h>   
struct point{ double x, y; };  
struct line{ point a, b; };  
  
double distance(point p1, point p2){  
    return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));  
}  
  
point intersection(line u, line v){  
    point ret = u.a;  
    double t = ((u.a.x - v.a.x)*(v.a.y - v.b.y) - (u.a.y - v.a.y)*(v.a.x - v.b.x))  
        / ((u.a.x - u.b.x)*(v.a.y - v.b.y) - (u.a.y - u.b.y)*(v.a.x - v.b.x));  
    ret.x += (u.b.x - u.a.x)*t;  
    ret.y += (u.b.y - u.a.y)*t;  
    return ret;  
}  
  
//外心   
point circumcenter(point a, point b, point c){  
    line u, v;  
    u.a.x = (a.x + b.x) / 2;  
    u.a.y = (a.y + b.y) / 2;  
    u.b.x = u.a.x - a.y + b.y;  
    u.b.y = u.a.y + a.x - b.x;  
    v.a.x = (a.x + c.x) / 2;  
    v.a.y = (a.y + c.y) / 2;  
    v.b.x = v.a.x - a.y + c.y;  
    v.b.y = v.a.y + a.x - c.x;  
    return intersection(u, v);  
}  
  
//內心   
point incenter(point a, point b, point c){  
    line u, v;  
    double m, n;  
    u.a = a;  
    m = atan2(b.y - a.y, b.x - a.x);  
    n = atan2(c.y - a.y, c.x - a.x);  
    u.b.x = u.a.x + cos((m + n) / 2);  
    u.b.y = u.a.y + sin((m + n) / 2);  
    v.a = b;  
    m = atan2(a.y - b.y, a.x - b.x);  
    n = atan2(c.y - b.y, c.x - b.x);  
    v.b.x = v.a.x + cos((m + n) / 2);  
    v.b.y = v.a.y + sin((m + n) / 2);  
    return intersection(u, v);  
}  
  
//垂心   
point perpencenter(point a, point b, point c){  
    line u, v;  
    u.a = c;  
    u.b.x = u.a.x - a.y + b.y;  
    u.b.y = u.a.y + a.x - b.x;  
    v.a = b;  
    v.b.x = v.a.x - a.y + c.y;  
    v.b.y = v.a.y + a.x - c.x;  
    return intersection(u, v);  
}  
  
//重心   
//到三角形三頂點距離的平方和最小的點   
//三角形內到三邊距離之積最大的點   
point barycenter(point a, point b, point c){  
    line u, v;  
    u.a.x = (a.x + b.x) / 2;  
    u.a.y = (a.y + b.y) / 2;  
    u.b = c;  
    v.a.x = (a.x + c.x) / 2;  
    v.a.y = (a.y + c.y) / 2;  
    v.b = b;  
    return intersection(u, v);  
}  
  
//費馬點   
//到三角形三頂點距離之和最小的點   
point fermentpoint(point a, point b, point c){  
    point u, v;  
    double step = fabs(a.x) + fabs(a.y) + fabs(b.x) + fabs(b.y) + fabs(c.x) + fabs(c.y);  
    int i, j, k;  
    u.x = (a.x + b.x + c.x) / 3;  
    u.y = (a.y + b.y + c.y) / 3;  
    while (step > 1e-10)  
        for (k = 0; k < 10; step /= 2, k++)  
            for (i = -1; i <= 1; i++)  
                for (j = -1; j <= 1; j++){  
                    v.x = u.x + step*i;  
                    v.y = u.y + step*j;  
                    if (distance(u, a) + distance(u, b) + distance(u, c) > distance(v, a) + distance(v, b) + distance(v, c))  
                        u = v;  
                }  
                return u;  
}  
int main()  
{  
    int n;  
    std::cin >> n;  
    while (n--)  
    {  
        point a, b, c;  
        std::cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y;  
        point center = perpencenter(a, b, c);  
        printf("%.4f %.4f\n", center.x, center.y);  
    }  
}  

#include <iostream>
#include <math.h>
#include <stdio.h>
struct point{ double x, y; };
struct line{ point a, b; };

double distance(point p1, point p2){
	return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
}

point intersection(line u, line v){
	point ret = u.a;
	double t = ((u.a.x - v.a.x)*(v.a.y - v.b.y) - (u.a.y - v.a.y)*(v.a.x - v.b.x))
		/ ((u.a.x - u.b.x)*(v.a.y - v.b.y) - (u.a.y - u.b.y)*(v.a.x - v.b.x));
	ret.x += (u.b.x - u.a.x)*t;
	ret.y += (u.b.y - u.a.y)*t;
	return ret;
}

//外心
point circumcenter(point a, point b, point c){
	line u, v;
	u.a.x = (a.x + b.x) / 2;
	u.a.y = (a.y + b.y) / 2;
	u.b.x = u.a.x - a.y + b.y;
	u.b.y = u.a.y + a.x - b.x;
	v.a.x = (a.x + c.x) / 2;
	v.a.y = (a.y + c.y) / 2;
	v.b.x = v.a.x - a.y + c.y;
	v.b.y = v.a.y + a.x - c.x;
	return intersection(u, v);
}

//內心
point incenter(point a, point b, point c){
	line u, v;
	double m, n;
	u.a = a;
	m = atan2(b.y - a.y, b.x - a.x);
	n = atan2(c.y - a.y, c.x - a.x);
	u.b.x = u.a.x + cos((m + n) / 2);
	u.b.y = u.a.y + sin((m + n) / 2);
	v.a = b;
	m = atan2(a.y - b.y, a.x - b.x);
	n = atan2(c.y - b.y, c.x - b.x);
	v.b.x = v.a.x + cos((m + n) / 2);
	v.b.y = v.a.y + sin((m + n) / 2);
	return intersection(u, v);
}

//垂心
point perpencenter(point a, point b, point c){
	line u, v;
	u.a = c;
	u.b.x = u.a.x - a.y + b.y;
	u.b.y = u.a.y + a.x - b.x;
	v.a = b;
	v.b.x = v.a.x - a.y + c.y;
	v.b.y = v.a.y + a.x - c.x;
	return intersection(u, v);
}

//重心
//到三角形三頂點距離的平方和最小的點
//三角形內到三邊距離之積最大的點
point barycenter(point a, point b, point c){
	line u, v;
	u.a.x = (a.x + b.x) / 2;
	u.a.y = (a.y + b.y) / 2;
	u.b = c;
	v.a.x = (a.x + c.x) / 2;
	v.a.y = (a.y + c.y) / 2;
	v.b = b;
	return intersection(u, v);
}

//費馬點
//到三角形三頂點距離之和最小的點
point fermentpoint(point a, point b, point c){
	point u, v;
	double step = fabs(a.x) + fabs(a.y) + fabs(b.x) + fabs(b.y) + fabs(c.x) + fabs(c.y);
	int i, j, k;
	u.x = (a.x + b.x + c.x) / 3;
	u.y = (a.y + b.y + c.y) / 3;
	while (step > 1e-10)
		for (k = 0; k < 10; step /= 2, k++)
			for (i = -1; i <= 1; i++)
				for (j = -1; j <= 1; j++){
					v.x = u.x + step*i;
					v.y = u.y + step*j;
					if (distance(u, a) + distance(u, b) + distance(u, c) > distance(v, a) + distance(v, b) + distance(v, c))
						u = v;
				}
				return u;
}
int main()
{
	int n;
	std::cin >> n;
	while (n--)
	{
		point a, b, c;
		std::cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y;
		point center = perpencenter(a, b, c);
		printf("%.4f %.4f\n", center.x, center.y);
	}
}