（HDU

（HDU - 3400）Line belt

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

Total Submission(s): 4640 Accepted Submission(s): 1796

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

题目大意：二维平面上有两条传送ab，cd，物体在传送带ab上的速度是p，在传送带cd上的速度是q，在其他地方的速度是r，问物体从a到d所需的时间最小是多少。

思路：最小的时间必然是在ab上走一段，cd上走一段，ab，cd之间走一段，设ab上的转折点为m，cd上的转折点为n，则时间t=am/p+mn/r+nd/q。假设m点确定，那么在cd之间三分即可求出点n使mn/r+nd/q最小，再在ab之间三分求整体时间最小，即三分嵌套。

#include<cstdio>#include<cmath>using namespace std;const double eps=1e-8;double p,q,r;struct node{    double x,y;};double dis(node a,node b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}node getmid(node a,node b){    node c;    c.x=(a.x+b.x)/2.0;    c.y=(a.y+b.y)/2.0;    return c;}double ts2(node m,node c,node d)//确定m在c,d之间ternary_search(三分){    node lo=c,hi=d,mid,midmid;    double t1=0,t2=1;//给t1,t2赋初值是为了防止刚开始t1，t2就很接近而不进行while循环     while(fabs(t1-t2)>eps)    {        mid=getmid(lo,hi);        midmid=getmid(mid,hi);        t1=dis(m,mid)/r+dis(mid,d)/q;        t2=dis(m,midmid)/r+dis(midmid,d)/q;        if(t1<t2) hi=midmid;        else lo=mid;     }     return t1;} double ts1(node a,node b,node c,node d)//再a,b之间整体三分 {    node lo=a,hi=b,mid,midmid;    double t1=0,t2=1;    while(fabs(t1-t2)>eps)    {        mid=getmid(lo,hi);        midmid=getmid(mid,hi);        t1=dis(a,mid)/p+ts2(mid,c,d);        t2=dis(a,midmid)/p+ts2(midmid,c,d);        if(t1<t2) hi=midmid;        else lo=mid;    }    return t1;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        node a,b,c,d;        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);        printf("%.2f\n",ts1(a,b,c,d));    }    return 0;}