三角形垂心

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转载最新推荐文章于 2022-02-26 09:23:45 发布 · 171 阅读

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原文链接：http://www.cnblogs.com/LitIce/archive/2011/06/06/2073627.html

给定一个三角形求垂心

跟前一篇求圆心的一样。
因为发现了我之前的做法都比较二，于是乎想把整个模板都改改。
我之前的模板的做法仍就很麻烦
我先一点和该点的对边求点斜式。然后两条直线求交点。
想法简单，但是代码长度比较长，

POJ1673

point orthocenter(point p0,point p1,point p2){
double a1,b1,a2,b2,c1,c2;
a1 = p2.x-p1.x; b1=p2.y-p1.y;c1 = 0;
a2 = p2.x-p0.x; b2=p2.y-p0.y;c2 = (p1.x-p0.x)*a2+(p1.y-p0.y)*b2;
double d = a1 * b2 - a2 * b1;
ret.x = p0.x + (c1 * b2 - c2 * b1) / d;
ret.y = p0.y + (a1 * c2 - a2 * c1) / d;
return ret;
}
读入三个点p0,p1,p2返回垂心
根据垂直向量点成为0
(O.x-p0.x)*(p2.x-p1.x)+(O.y-p0.y)*(p2.y-p1.y) = 0; (1)
(O.x-p1.x)*(p2.x-p0.x)+(O.y-p1.y)*(p2.y-p0.y) = 0; (2)

令O.x = x+p0.x;O.y=y+p0.y带入(1)(2)得到
x*(p2.x-p1.x)+y*(p2.y-p1.y) = 0;
x*(p2.x-p0.x)+y*(p2.y-p0.y) = (p2.x-p0.x)*(p1.x-p0.x)+(p2.y-p0.y)*(p1.y-p0.y);
解得(x,y)
O = (x,y)+p0;

#include<iostream>
#include<cstdio>

using namespace std;

struct point{
	double x,y;
} p0,p1,p2,ret;
int Test;

point orthocenter(point p0,point p1,point p2){
	double a1,b1,a2,b2,c1,c2;
	a1 = p2.x-p1.x; b1=p2.y-p1.y;c1 = 0;
	a2 = p2.x-p0.x; b2=p2.y-p0.y;c2 = (p1.x-p0.x)*a2+(p1.y-p0.y)*b2;
	double d = a1 * b2 - a2 * b1; 
 	ret.x = p0.x + (c1 * b2 - c2 * b1) / d; 
 	ret.y = p0.y + (a1 * c2  - a2 * c1) / d; 
	return ret;
}
int main(){
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
	cin >> Test;
	for (int ii = 1;ii<=Test;++ii){
		cin>>p0.x>>p0.y>>p1.x>>p1.y>>p2.x>>p2.y;
		ret = orthocenter(p0,p1,p2);
	     printf("%.4lf %.4lf\n",ret.x,ret.y);
	}
	return 0;
}

转载于:https://www.cnblogs.com/LitIce/archive/2011/06/06/2073627.html