hdu 3400 Line belt(多重三分)

题目：​​http://acm.hdu.edu.cn/showproblem.php?pid=3400​​

​​Line belt​​

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

​​Submit​​​ ​​Status​​

Description

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

分析：最初自己想到的是向量，三个主要的向量：AB上的a向量，CD上的c向量，其他区域的b向量。三者满足这样的关系：(1)|a+b+c|=|AD|  (2)|a|/p+|b|/r+|c|/q=min_time.  (3) 0<=|a|<=|AB|   (4)0<=|c|<=|CD|   (5) |b|=|AD|-|a+c|  因为是极值问题自然想到三分，可是向量怎么用长度表示？不能，只有用二维坐标，把所有的位置信息存在点里，计算距离，第三个点的位置推导全部用坐标来解决。

pt,qt都是两条线段上运动的点，pt,qt的坐标可以通过线段长度之比推导出来。最后的结果确定：每定一个pt点就对qt三分，找A--pt,pt--qt,qt--D的长度之和，pt用一次三分，qt在pt的基础上再用三分，即pt在AB上动一下，qt在CD上转悠一圈。

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const double eps=1e-7,Min=1e-9;
struct node{
    double x,y;
}a,b,c,d,pt,qt;
double lab,lcd,p,q,r;
double dist(node q1,node q2){
    return sqrt(eps+(q1.x-q2.x)*(q1.x-q2.x)+(q1.y-q2.y)*(q1.y-q2.y));  //eps不能缺
}
void calpt(double ra){
    pt.x=a.x+ra/lab*(b.x-a.x);
    pt.y=a.y+ra/lab*(b.y-a.y);
}
void calqt(double ra){
    qt.x=d.x+ra/lcd*(c.x-d.x);
    qt.y=d.y+ra/lcd*(c.y-d.y);
}
double cal(double ra){
     calpt(ra);
     double low=0,high=lcd,mid1,mid2,ans=dist(pt,a)/p,ans2,ans3;
     while(high-low>Min){
         mid1=low+(high-low)/3;
         mid2=high-(high-low)/3;
         calqt(mid1);
         ans2=dist(qt,pt)/r+dist(qt,d)/q;
         calqt(mid2);
         ans3=dist(qt,pt)/r+dist(qt,d)/q;
         if(ans2<ans3) high=mid2;
         else low=mid1;
     }
     return ans+ans2;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int t;
    cin>>t;
    while(t--){
        double ax,ay,bx,by,cx,cy,dx,dy;
        scanf("%lf%lf%lf%lf",&ax,&ay,&bx,&by);
        a.x=ax;  a.y=ay;   b.x=bx;   b.y=by;
        scanf("%lf%lf%lf%lf",&cx,&cy,&dx,&dy);
        c.x=cx;  c.y=cy;   d.x=dx;   d.y=dy;
        scanf("%lf%lf%lf",&p,&q,&r);
        lab=dist(a,b),lcd=dist(c,d);
        double low=0,high=lab,mid1,mid2;
        while(high-low>Min){
             mid1=low+(high-low)/3;
             mid2=high-(high-low)/3;
             if(cal(mid1)<cal(mid2))  high=mid2;  //边界不要混淆
             else  low=mid1;
        }
        printf("%.2lf\n",cal(mid1));
    }
    return 0;
}