HDU 3400 Line belt （三分嵌套）

题目链接

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 }

 

posted @ 2014-11-24 21:47  水门  阅读(432)  评论(0)  编辑  收藏  举报