Line belt 三分嵌套

 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

因为有两个变量所以用两个三分
用cin会tle。。。没想到会在这超时。。。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
struct point
{
    double x,y;
}a,b,c,d;
double P,Q,R;
double dis(point m,point n)
{
    return sqrt((m.x-n.x)*(m.x-n.x)+(m.y-n.y)*(m.y-n.y));
}
double calcd(point m)
{
    double t1,t2;
    point m1,m2,l=c,r=d;
    do
    {
        m1.x=(r.x+l.x)/2;
        m1.y=(r.y+l.y)/2;
        m2.y=(m1.y+r.y)/2;
        m2.x=(m1.x+r.x)/2;
        t1=dis(d,m1)/Q+dis(m,m1)/R;
        t2=dis(d,m2)/Q+dis(m,m2)/R;
        if(t1>t2)
            l=m1;
        else
            r=m2;
    }while(dis(m1,m2)>=1e-4);
    return t1;
}
double calab()
{
    double t1,t2;
    point m1,m2,l=a,r=b;
    do
    {
        m1.x=(r.x+l.x)/2;
        m1.y=(r.y+l.y)/2;
        m2.y=(m1.y+r.y)/2;
        m2.x=(m1.x+r.x)/2;
        t1=dis(a,m1)/P+calcd(m1);
        t2=dis(a,m2)/P+calcd(m2);
        if(t1>t2)
            l=m1;
        else
            r=m2;
        //cout<<m1.x<<' '<<m1.y<<endl;
        //cout<<m2.x<<' '<<m2.y<<endl;
    }while(dis(m1,m2)>=1e-4);
    return t1;
}
int main()
{
    int t;
    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);
        double ans=calab();
        printf("%.2lf
",ans);
    }
    return 0;
}