hdu 3400 Line belt

Line belt

                                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                              Total Submission(s): 2973    Accepted Submission(s): 1139

Problem 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

 

Author

lxhgww&&momodi

 

题意：

        已知两条线段，分别知道在两条线段上AB,CD的速度p，q和不在两条线段上的速度r，求从A点到D点的最小时间。

题解：

         三分题，时间最短的路径必定是至多3条直线段构成的，一条在AB上，一条在CD上，一条在两条线段之间通过三分AB间的位置，确定在AB的位置，后三分CD的位置找到最小值，这样两个三分嵌套在一起就可以了。

        至于为什么用三分，可以想下直接从A到D距离最短，但是有可能因为速度比较慢而耽误时间，可以在AB，CD上走一段来缩小时间，但是在AB，CD上走得过于多就会因为路程增加而导致时间增加，所以是单峰函数。

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
const double eps=1e-6;
struct node
{
    double x;
    double y;
}a,b,c,d,bb,dd;
double ab,cd;
double p,q,r;
double dis(node u,node v)
{
    return  sqrt(eps+(u.x-v.x)*(u.x-v.x)+(u.y-v.y)*(u.y-v.y));
}
double work(double t)
{
    dd.x=d.x+(c.x-d.x)/cd*t*q;
    dd.y=d.y+(c.y-d.y)/cd*t*q;
    return t+dis(bb,dd)/r;
}
double solve(double t)
{
     bb.x=a.x+(b.x-a.x)/ab*t*p;
     bb.y=a.y+(b.y-a.y)/ab*t*p;
    double l=0,h=cd/q,mid1,mid2;
    while(l+eps