搜索(三分)：HDU 3400 Line belt

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3531    Accepted Submission(s): 1364

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

 

　　

 //rp++
 //#include <bits/stdc++.h>

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <cmath>

 using namespace std;

 double eps=1e-;

 double Ax,Ay,Bx,By,Cx,Cy,Dx,Dy,P,Q,R;

 double DIS(double x1,double y1,double x2,double y2)
 {
     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
 }

 double Get_Time(double p1,double p2)
 {
     return DIS(Ax,Ay,Ax+p1*(Bx-Ax),Ay+p1*(By-Ay))*1.0/P+DIS(Dx,Dy,Dx+p2*(Cx-Dx),Dy+p2*(Cy-Dy))*1.0/Q+
            DIS(Bx+(1.0-p1)*(Ax-Bx),By+(1.0-p1)*(Ay-By),Cx+(1.0-p2)*(Dx-Cx),Cy+(1.0-p2)*(Dy-Cy))*1.0/R;
 }

 double Get_MIN(double pos)
 {
     double L=0.0,R=1.0,x,y,ret;
     while(R-L>=eps)
     {
         x=(R+*L)/3.0;
         y=(*R+L)/3.0;

         x=Get_Time(pos,x);
         y=Get_Time(pos,y);

         if(x-y>=eps)L=(R+*L)/3.0;
         else R=(*R+L)/3.0;

         ret=min(x,y);
     }
     return ret;
 }

 double Solve()
 {
     double L=0.0,R=1.0,x,y,ret;
     while(R-L>=eps)
     {
         x=(*L+R)/3.0;
         y=(L+*R)/3.0;

         x=Get_MIN(x);
         y=Get_MIN(y);

         if(x-y>=eps)L=(*L+R)/3.0;
         else R=(L+*R)/3.0;

         ret=min(x,y);
     }
     return ret;
 }

 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%lf%lf%lf%lf",&Ax,&Ay,&Bx,&By);
         scanf("%lf%lf%lf%lf",&Cx,&Cy,&Dx,&Dy);
         scanf("%lf%lf%lf",&P,&Q,&R);

         printf("%.2lf\n",Solve());
     }
     return ;
 }