搜索(三分)：HDU 3400 Line belt

Posted 2020-06-21 既然选择了远方，便只顾风雨兼程

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

 

　　

 1 //rp++
 2 //#include <bits/stdc++.h>
 3 
 4 #include <iostream>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <cmath>
 8 
 9 using namespace std;
10 
11 double eps=1e-8;
12 
13 double Ax,Ay,Bx,By,Cx,Cy,Dx,Dy,P,Q,R;
14 
15 double DIS(double x1,double y1,double x2,double y2)
16 {
17     return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
18 }
19 
20 
21 double Get_Time(double p1,double p2)
22 {
23     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+
24            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;
25 }
26 
27 double Get_MIN(double pos)
28 {
29     double L=0.0,R=1.0,x,y,ret;
30     while(R-L>=eps)
31     {
32         x=(R+2*L)/3.0;
33         y=(2*R+L)/3.0;
34         
35         x=Get_Time(pos,x);
36         y=Get_Time(pos,y);
37         
38         
39         if(x-y>=eps)L=(R+2*L)/3.0;
40         else R=(2*R+L)/3.0;
41         
42         ret=min(x,y);
43     }
44     return ret;
45 }
46 
47 double Solve()
48 {
49     double L=0.0,R=1.0,x,y,ret;
50     while(R-L>=eps)
51     {
52         x=(2*L+R)/3.0;
53         y=(L+2*R)/3.0;
54         
55         x=Get_MIN(x);
56         y=Get_MIN(y);
57         
58         
59         if(x-y>=eps)L=(2*L+R)/3.0;
60         else R=(L+2*R)/3.0;
61         
62         ret=min(x,y);
63     }
64     return ret;
65 }
66 
67 int main()
68 {
69     int T;
70     scanf("%d",&T);
71     while(T--)
72     {
73         scanf("%lf%lf%lf%lf",&Ax,&Ay,&Bx,&By);
74         scanf("%lf%lf%lf%lf",&Cx,&Cy,&Dx,&Dy);
75         scanf("%lf%lf%lf",&P,&Q,&R);
76         
77         printf("%.2lf\n",Solve());
78     }
79     return 0;
80 }