Line belt（三分镶嵌）

In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane.
How long must he take to travel from A to D?

Input

The first line is the case number T.
For each case, there are three lines.
The first line, four integers, the coordinates of A and B: Ax Ay Bx By.
The second line , four integers, the coordinates of C and D:Cx Cy Dx Dy.
The third line, three integers, P Q R.
0<= Ax，Ay，Bx，By，Cx，Cy，Dx，Dy<=1000
1<=P，Q，R<=10

 

Output

The minimum time to travel from A to D, round to two decimals.

 

Sample Input

1
0 0 0 100
100 0 100 100
2 2 1

Sample Output

136.60


题目意思：就是给你两条线段AB , CD的坐标 ，一个人在AB以速度p跑,在CD上以q跑,在其他地方跑速度是r，问你从A到D最少的时间。
解题思路：设E在AB上，F在CD上。 则人在线段AB上花的时间为：f = AE / p，人走完Z和Y所花的时间为：g= EF / r + FD / q。 
f函数是一个单调递增的函数，而g很明显是一个先递减后递增的函数。两个函数叠加，所得的函数应该也是一个先递减后递增的函数。这算是一道三分又三分的题目，可以看成是三分的镶嵌,
先对AB上的位置E进行三分枚举，每一次枚举的时候把E看做定点，在这种情况下，再对CD上的位置F进行三分枚举，这样可以求出此次三分AB的最小时间。然后完成对AB的三分后，就会得到从A到D的最短时间。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
#define  EPS 1e-8
struct Point
{
    double x;
    double y;
} a, b, c, d, e, f;
double p, q, r;
double dis(Point p1, Point p2)///两点之间的距离
{
    return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
}
double calc(double alpha)///alpha代表在f点在cd中位置的比率
{
    f.x = c.x + (d.x - c.x) * alpha;
    f.y = c.y + (d.y - c.y) * alpha;
    return dis(f, d) / q + dis(e, f) / r;///返回在ef和fd上花费的时间
}
double inter_tri(double alpha)///在f点进行三分
{
    double l = 0.0, r = 1.0, mid, mmid, cost;
    e.x = a.x + (b.x - a.x) * alpha;
    e.y = a.y + (b.y - a.y) * alpha;///e点在线段ab中的位置
    while (r - l > EPS)
    {
        mid = (l + r) / 2;
        mmid = (mid + r) / 2;
        cost = calc(mid);
        if (cost <= calc(mmid))
            r = mmid;
        else
            l = mid;
    }
    return dis(a, e) / p + cost;
}
double solve()///在e点进行三分
{
    double l = 0.0, r = 1.0, mid, mmid, ret;
    while (r - l > EPS)
    {
        mid = (l + r) / 2;
        mmid = (mid + r) / 2;
        ret = inter_tri(mid);
        if (ret <= inter_tri(mmid))
            r = mmid;
        else
            l = mid;
    }
    return ret;
}                       
int main()
{
    int T;
    double ans;
    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("%lf%lf%lf",&p,&q,&r);
        ans=solve();
        printf("%.2lf\n",ans);
    }
    return 0;
}