acm 2 1006 Line belt

1.1006

2.

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

3.在二维平面上有两线带,有两段AB和CD,lxhgww的速度对AB是P,在CD是Q,他用速度R在其他地区在平面上移动。多久他从A到D?

4.先三分AB上的点,再三分CD上的点
证明: 
设E在AB上,F在CD上。 
AB 时间:f = AE / p,
Z和Y 时间:g = EF / r + FD / q。 
f单调递增,而g先递减后递增。
两个函数叠加,所得的函数先递减后递增,所以用三分法

5.#include
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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;
    cin>>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);
        cin>>p>>q>>r;
        printf("%.2f\n",find(a,b,c,d));
    }
    return 0;
}

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