Line belt

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

题意：有两个传送带，由A到B的速度是p，有C到D的速度是q，其他速度是r，求由A到D的最少时间；如图（原创）

belt" title="Line belt">

  

解题思路：在A，B之间去mid，然后三分求最小时间，三分过程中是A到mid的时间加上mid到D的时间（mid到D的时间，也用三分求，最好写一个函数，求A到mid的时间用三分时，每计算一次，都得在当前状态求一个mid到D的时间），然后就能得出来最小时间；

感悟：思路不好想；刚开始想的是，在AB中去mid先求出最合适的mid到D的时间，然后把B转移到mid，然后再在CD上取mid，求最合适的mid；这样不是动态中求得，不行。

思路不好写，代码更不好写啊，用do，while能过，用while就过不了；

代码：

#include

#include

#include

#include

#include

using namespace std;

const double eps=1e-6;





struct coordinate

{

    double
x;

    double
y;

};

coordinate A,B,C,D;

int t;

double p,q,r;

double dis(coordinate a,coordinate b)

{

    double
t;

   
t=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

   
//printf("t=%.2f\n",t);

    return
t;

}//求两点之间的距离

coordinate midc(coordinate a,coordinate b)

{

    coordinate
t;

   
t.x=(a.x+b.x)*0.5;

   
t.y=(a.y+b.y)*0.5;

    return
t;

}//求坐标的中点

double time(coordinate a,coordinate b,coordinate c,coordinate
d)

{

    double
t;

   
t=dis(a,b)/p+dis(b,c)/r+dis(c,d)/q;

   
//printf("t=%.2f\n",t);

    return
t;

}//求路径的总时间

double Three_algorithm_1(coordinate a,coordinate c,coordinate
d)

{

    double
t1,t2;

    coordinate
left,right,mid,midmid;

   
left=c;

   
right=d;

    do

    {

       
mid=midc(left,right);

       
midmid=midc(mid,right);

      
// printf("dis(right,left)=%.5f\n",dis(right,left));

       
t1=dis(a,mid)/r+dis(mid,d)/q;

       
t2=dis(a,midmid)/r+dis(midmid,d)/q;

       
if(t1>t2)left=mid;

       
else right=midmid;

   
}while(dis(right,left)>=eps);

    return
t1;

}//先求出A到CD段的最小时间

double Three_algorithm_2(coordinate a,coordinate b,coordinate
c,coordinate d)

{

    double
t1,t2;

    coordinate
left,right,mid,midmid;

   
left=a;

   
right=b;

   
mid=midc(left,right);

   
midmid=midc(mid,right);

    do

    {

       
mid=midc(left,right);

       
midmid=midc(mid,right);

       
//printf("dis(right,left)=%.f\n",dis(right,left));

       
t1=dis(a,mid)/p+Three_algorithm_1(mid,c,d);

       
t2=dis(a,midmid)/p+Three_algorithm_1(midmid,c,d);

       
if(t1>t2)left=mid;

       
else right=midmid;

   
}while(dis(right,left)>=eps);

    return
t1;

}//加上前面求出的先求出A到CD段的最小时间，用三分求最少时间

int main()

{

   
//freopen("in.txt", "r", stdin);

   
scanf("%d",&t);

    for(int
i=0;i

    {

       
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("%.2lf\n",Three_algorithm_2(A,B,C,D));

    }

}