HDU3400(计算几何中的三分法利用)

题目：Line belt

 

题意：就是给你两条线段AB , CD ，一个人在AB上跑速度p, 在CD上跑q,在其他地方跑速度是r。问你从A到D最少的时间是好多。

#include<iostream>
#include<cmath>
#include <stdio.h>
using namespace std;

typedef struct
{
	double x , y ;
}point;

point a,b,c,d;
int  p , q , r ;
const double eps = 1e-8;

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

double part(double k)
{
	double Lx = 0 , Rx = dis(c, d);
	point e , f1 ,f2;
	for(int i = 0 ; i < 200 ;++i)
	{
		double len = Rx - Lx ;
		double t1 = Lx  + len * 4/9;
		double t2 = Lx  + len *5/9;
		e.x = a.x + k * (b.x - a.x );
		e.y = a.y + k * ( b.y -a.y );
		f1.x = c.x + (t1 * (d.x - c.x ));
		f1.y = c.y + (t1* (d.y - c.y ));
		f2.x = c.x + (t2 * (d.x - c.x ));
		f2.y = c.y + (t2 * (d.y - c.y ));
		double res1 = dis(a,e)/p + dis(e,f1)/r + dis(f1,d)/q;
		double res2 = dis(a,e)/p + dis(e,f2)/r + dis(f2,d)/q;
		if( res1 < res2) Rx = t2 ;
		else Lx = t1 ;
	}
	double t = (Lx+Rx)/2.0;
	e.x = a.x + (k * (b.x - a.x ))  ;
	e.y = a.y + (k * ( b.y -a.y )) ;
	f1.x = c.x + (t * (d.x - c.x ));
	f1.y = c.y + (t* (d.y - c.y));
	return  dis(a,e)/p + dis(e,f1)/r + dis(f1,d)/q;
}

double solve(double Min , double Max )
{
	double Lx = Min , Rx = Max ;
	for(int i = 0 ; i < 200 ; ++i)
	{
		double len = Rx - Lx ;
		double t1 = Lx + len * 4/9 ;
		double t2 = Lx  + len* 5/9 ;
		double f1 = part(t1);
		double f2 = part(t2);
		if( f1 < f2) Rx = t2 ;
		else    Lx = t1;
	}
	return part((Lx+Rx)/2.0);
}
int main()
{
	int t ;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
		scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);
		scanf("%d%d%d",&p,&q,&r);
		double res = solve(0.0, 1.0);
		printf("%.2f\n",res);
	}
	return 0 ;
}