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  • poj1673

    所谓Exocenter就是垂心。不难证明。


    #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);
    }
    
    //内心
    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
    //到三角形三顶点距离的平方和最小的点
    //三角形内到三边距离之积最大的点
    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);
    }
    
    //费马点
    //到三角形三顶点距离之和最小的点
    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
    ", center.x, center.y);
    	}
    }


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  • 原文地址:https://www.cnblogs.com/javawebsoa/p/3230766.html
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