Line belt
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2862 Accepted Submission(s): 1099
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
Author
lxhgww&&momodi
题意：
给出两条传送带的起点到末端的坐标，其中ab为p的速度，cd为q的速度 其他地方为r的速度
求a到d点的最短时间。
分析：
首先要看出来这是一个凹型的函数，
时间最短的路径必定是至多3条直线段构成的，一条在AB上，一条在CD上，一条架在两条线段之间。
所有利用两次三分，第一个三分ab段的一点，第二个三分知道ab一点后的cd段的接点。
刚开始没用do while错了两次，因为如果给的很接近的话，上来的t1没有赋值。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cstdlib>
5 #include <cmath>
6 #include <algorithm>
7 #define LL __int64
8 const int maxn = 1e2 + 10;
9 const double eps = 1e-8;
10 using namespace std;
11 double p, q, r;
12 struct node
13 {
14 double x, y;
15 }a, b, c, d;
16
17 double dis(node a, node b)
18 {
19 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
20 }
21
22 double solve2(node t)
23 {
24 double d1, d2;
25 node le = c;
26 node ri = d;
27 node mid, midmid;
28 do
29 {
30 mid.x = (le.x+ri.x)/2.0;
31 mid.y = (le.y+ri.y)/2.0;
32 midmid.x = (mid.x+ri.x)/2.0;
33 midmid.y = (mid.y+ri.y)/2.0;
34 d1 = dis(t, mid)/r + dis(mid, d)/q;
35 d2 = dis(t, midmid)/r + dis(midmid, d)/q;
36 if(d1 > d2)
37 le = mid;
38 else ri = midmid;
39 }while(dis(le, ri)>=eps);
40 return d1;
41 }
42
43 double solve1()
44 {
45 double d1, d2;
46 node le = a;
47 node ri = b;
48 node mid, midmid;
49 do
50 {
51 mid.x = (le.x+ri.x)/2.0;
52 mid.y = (le.y+ri.y)/2.0;
53 midmid.x = (mid.x+ri.x)/2.0;
54 midmid.y = (mid.y+ri.y)/2.0;
55 d1 = dis(a, mid)/p + solve2(mid);
56 d2 = dis(a, midmid)/p + solve2(midmid);
57 if(d1 > d2)
58 le = mid;
59 else ri = midmid;
60 }while(dis(le, ri)>=eps);
61 return d1;
62 }
63
64 int main()
65 {
66 int t;
67 scanf("%d", &t);
68 while(t--)
69 {
70 scanf("%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y);
71 scanf("%lf%lf%lf%lf", &c.x, &c.y, &d.x, &d.y);
72 scanf("%lf%lf%lf", &p, &q, &r);
73 printf("%.2lf\n", solve1());
74 }
75 return 0;
76 }