CSU-ACM2017暑期训练5-三分 E - Line belt HDU - 3400 

题目：

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

题意：给你两条平面坐标中的线段AB,CD的坐标.有一人想从A点走到D点，已知这人在AB上的速度p，在CD上的速度q和在坐标系其他地方的速度r.求A到达D的最短时间.

思路：既然要求时间短，肯定是在AB上找一点E从A走到E然后在CD上找一点F又从E走到F，然后走FD.那么问题是如何确定这个点.根据总时间s=p/AE+r/EF+q/fD可得.于是我们分别对AB和CD进行三分来寻找满足要求的最小的点E和F.也就是嵌套三分.具体见代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn = 100;

struct Point
{
    double x,y;
}a,b,c,d,e,f,g;

double p,q,r;
int T,n,m;

double get_dis(Point v,Point u)
{
    return sqrt((v.x-u.x)*(v.x-u.x)+(v.y-u.y)*(v.y-u.y));
}

double cal(double rat_cd)//计算cd以及其余路上的长度
{
    f.x=c.x+(d.x-c.x)*rat_cd;
    f.y=c.y+(d.y-c.y)*rat_cd;
    return get_dis(f,d)/q+get_dis(e,f)/r;
}

double ts_2(double rat_ab)//三分cd中f点的位置
{
    e.x=a.x+(b.x-a.x)*rat_ab;
    e.y=a.y+(b.y-a.y)*rat_ab;
    double l=0,r=1.0;
    while(r-l>eps)
    {
        double mid,midmid;
        mid=(l+r)/2;
        midmid=(mid+r)/2;
        if(cal(mid)>cal(midmid))
            l=mid;
        else
            r=midmid;
    }
    return get_dis(a,e)/p+cal(l);
}

double ts_1()//三分ab中的e点所在位置
{
    double l=0,r=1.0;
    while(r-l>eps)
    {
        double mid,midmid;
        mid=(l+r)/2;
        midmid=(mid+r)/2;
        if(ts_2(mid)>ts_2(midmid))
            l=mid;
        else
            r=midmid;
    }
    return ts_2(l);
}

int main()
{
    scanf("%d",&T);
	while(T--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y);
        scanf("%lf%lf%lf",&p,&q,&r);
        printf("%.2f\n",ts_1());
    }
    return 0;
}