hdu Line belt

Line belt

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 85 Accepted Submission(s): 47

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

 三分搜索：

如图所示必定在AB CD上存在ＸＹ 使得A---D 用时间最短。用三分法针对AB间的某个点找出在CD段上最小的点，

嵌套循环找出最小的时间。有关时间函数在AB段和在CD段均为凸函数，所以可以用三分法。

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
 struct point
{
    double x,y;
};
int p,q,r,v;
double dis(point a,point b)
{
    return pow(pow(a.y-b.y,2)+pow(a.x-b.x,2),1.0/2);
}
double findy(point c,point d,point y)
{
    point mid,midmid,left,right;
    double t1,t2;
    left = c;
    right = d;
    do
    {
        mid.x = (left.x+right.x)/2;
        mid.y = (left.y+right.y)/2;
        midmid.x = (right.x+mid.x)/2;
        midmid.y = (right.y+mid.y)/2;
        t1 = dis(d,mid)/q+dis(mid,y)/r;
        t2 = dis(d,midmid)/q+dis(midmid,y)/r;
        if(t1>t2)
        left = mid;
        else right = midmid;
    }while(fabs(t1-t2)>0.000001);
    return t1;
}
double find(point a,point b,point c,point d)
{
    point mid,midmid,left,right;
    double t1,t2;
    left = a;
    right = b;
    do
    {
        mid.x = (left.x+right.x)/2;
        mid.y = (left.y+right.y)/2;
        midmid.x = (right.x+mid.x)/2;
        midmid.y = (right.y+mid.y)/2;
        t1 = dis(a,mid)/p+findy(c,d,mid);
        t2 = dis(a,midmid)/p+findy(c,d,midmid);
        if(t1>t2)left = mid;
        else right = midmid;
    }while(fabs(t1-t2)>0.000001);
    return t1;
}
int main()
{
    int t;
    point a,b,c,d;
    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("%d%d%d",&p,&q,&r);
        printf("%.2f\n",find(a,b,c,d));
    }
    return 0;
}