题目大意：给出平面上两条线段，在这两条线段上走有一定的速度，在其他的平面上走也有一定的速度，问从A点到D点最少需要多少时间。

思路：好像是三分吧，大概感受一下吧，反正也不会证。

CODE：

#include <cmath>
#include <cstdio>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define EPS 1e-7
#define INF 1e15
using namespace std;

struct Point{
	double x,y;
	
	Point(double _,double __):x(_),y(__) {}
	Point() {}
	void Read() {
		scanf("%lf%lf",&x,&y);
	}
	Point operator +(const Point &a)const {
		return Point(x + a.x,y + a.y);
	}
	Point operator -(const Point &a)const {
		return Point(x - a.x,y - a.y);
	}
	Point operator *(double a)const {
		return Point(x * a,y * a);
	}
	Point operator /(double a)const {
		return Point(x / a,y / a);
	}
}A,B,C,D;

inline double Calc(const Point &a,const Point &b)
{
	return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

double v1,v2,v3;

double TriDivide(const Point &p,Point l,Point r)
{
	while(Calc(l,r) > EPS) {
		Point u = (r - l) / 3.0;
		Point p1 = l + u,p2 = p1 + u;
		double ans1 = Calc(A,p) / v1 + Calc(p1,D) / v2 + Calc(p,p1) / v3;
		double ans2 = Calc(A,p) / v1 + Calc(p2,D) / v2 + Calc(p,p2) / v3;
		if(ans1 > ans2)	l = p1;
		else	r = p2;
	}
	return Calc(A,p) / v1 + Calc(l,D) / v2 + Calc(p,l) / v3;
}

double TriDivide(Point l,Point r)
{
	while(Calc(l,r) > EPS) {
		Point u = (r - l) / 3.0;
		Point p1 = l + u,p2 = p1 + u;
		double ans1 = TriDivide(p1,C,D),ans2 = TriDivide(p2,C,D);
		if(ans1 > ans2)	l = p1;
		else	r = p2;
	}
	return TriDivide(l,C,D);
}

int main()
{
	A.Read(),B.Read(),C.Read(),D.Read();
	scanf("%lf%lf%lf",&v1,&v2,&v3);
	cout << fixed << setprecision(2) << TriDivide(A,B) << endl;
	return 0;
}

BZOJ 1857 SCOI 2010 传送带 三分法