UVA 11178 - Morley's Theorem 向量

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2119

题目大意：

Morley定理是这样定义的，做三角形ABC每个内角的三等分线，相交成三角形DEF，则DEF是等边三角形。如图，你的任务是根据A,B,C三个点的位置确定D、E、F的位置。

思路：

把BC边旋转三分之一的角ABC，CB边旋转三分之一的角ACB，然后求交点D就出来了。其他的同理。

PS：作者的向量类写得不错，嗯，等我回学校了要把c++ primer plus里的向量模版好好看看

#include<cstdio>

#include<cmath>

struct point

{

	double x,y;

	point(double x=0,double y=0): x(x),y(y){}

};

typedef point Vector;

Vector operator +(point a,point b)

{

	return Vector(a.x+b.x,a.y+b.y);

}

Vector operator *(point a,double b)

{

	return Vector(a.x*b,a.y*b);

}

Vector operator -(point a,point b)

{

	return Vector(a.x-b.x,a.y-b.y);

}

double dot(Vector a,Vector b)

{

	return a.x*b.x+a.y*b.y;

}

double cross(Vector a,Vector b)

{

	return a.x*b.y-a.y*b.x;

}

double len(Vector a)

{

	return sqrt(a.x*a.x+a.y*a.y);

}

Vector rotate(Vector a,double rad)

{

	return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));

}

point getans(point p,Vector v,point q,Vector w){

	Vector u= p-q;

	double t=cross(w,u) / cross(v,w);

	return p+v*t;

}

point getPoint(point a,point b,point c)

{

	Vector bc=c-b;

	Vector ba=a-b;

	double x=acos(  dot(ba,bc) / len(bc) / len(ba) );

	Vector bd= rotate(bc,x/3);

	Vector ca=a-c;

	Vector cb=b-c;

	x=acos(  dot(cb,ca) / len(cb) / len(ca) );

	Vector cd=rotate(cb,-x/3);

	return getans(b,bd,c,cd);

}

int main()

{

	int T;

	scanf("%d",&T);

	point a,b,c,d,e,f;

	while(T--)

	{

		scanf("%lf%lf",&a.x,&a.y);

		scanf("%lf%lf",&b.x,&b.y);

		scanf("%lf%lf",&c.x,&c.y);

		d=getPoint(a,b,c);

		e=getPoint(b,c,a);

		f=getPoint(c,a,b);

		printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n",d.x,d.y,e.x,e.y,f.x,f.y);

	}

	return 0;

}

巴特西

UVA 11178 - Morley's Theorem 向量

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