hdu 3400 Line belt

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3023    Accepted Submission(s): 1155

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

10 0 0 100100 0 100 1002 2 1

 

Sample Output

136.60

 

Author

lxhgww&&momodi

 

1. 分析：基本思想就时枚举线段AB，CD上所有点的组合，取最优解，在次基础上
2.         用三分优化，至于为何用三分，是因为它是凸函数，在此不予证明。
3. 献上代码：
4.  #include<stdio.h>
5. #include<string.h>
6. #include<algorithm>
7. #include<iostream>
8. #include<math.h>
9. #define eps 1e-9
10. using namespace std;
11. double ax,ay,bx,by,cx,cy,dx,dy;
12. double k1,l1,l2;
13. double p,q,r1;
14. double dis(double ax,double ay,double bx,double by)
15. {
16.     return sqrt((bx-ax)*(bx-ax)+(by-ay)*(by-ay));
17. }
18. double cal(double k2)
19. {
20.     double t1,t2,t3;
21.     t1=l1*k1/p;
22.     t2=l2*(1-k2)/q;
23.     t3=dis(ax+(bx-ax)*k1,ay+(by-ay)*k1,cx+(dx-cx)*k2,cy+(dy-cy)*k2)/r1;
24.     return t1+t2+t3;
25. }   
26. double thr()
27. {
28.     double l,r,mid,midmid,dis1,dis2;
29.     l=0,r=1;
30.     while(r-l>=eps)
31.     {   
32.         mid=(l+r)/2;
33.         midmid=(mid+r)/2;
34.         dis1=cal(mid);
35.         dis2=cal(midmid);
36.         if(dis1<dis2)
37.         {
38.             r=midmid;
39.         }
40.         else
41.         {
42.             l=mid;
43.         }
44.     }
45.     return dis1<dis2?dis1:dis2;
46. }
47. int main()
48. {
49.     int T;
50.     double l,r,mid,midmid,dis1,dis2;
51.     scanf("%d",&T);
52.     while(T--)
53.     {
54.         scanf("%lf%lf%lf%lf",&ax,&ay,&bx,&by);
55.         scanf("%lf%lf%lf%lf",&cx,&cy,&dx,&dy);
56.         scanf("%lf%lf%lf",&p,&q,&r1);
57.         l=0,r=1;
58.         l1=dis(ax,ay,bx,by);
59.         l2=dis(cx,cy,dx,dy);
60.         while(r-l>=eps)
61.         {   
62.             mid=(l+r)/2;
63.             midmid=(mid+r)/2;
64.             k1=mid;
65.             dis1=thr();
66.             k1=midmid;
67.             dis2=thr();
68.             if(dis1<dis2)
69.             {
70.                 r=midmid;
71.             }
72.             else
73.             {
74.                 l=mid;
75.             }
76.         }
77.         printf("%.2lf\n",dis1<dis2?dis1:dis2);
78.     }
79.     return 0;
80. }
81.         
82.     
83.