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。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <algorithm>
 #define LL __int64
 const int maxn = 1e2 + ;
 const double eps = 1e-;
 using namespace std;
 double p, q, r;
 struct node
 {
     double x, y;
 }a, b, c, d;

 double dis(node a, node b)
 {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }

 double solve2(node t)
 {
     double d1, d2;
     node le = c;
     node ri = d;
     node mid, midmid;
     do
     {
         mid.x = (le.x+ri.x)/2.0;
         mid.y = (le.y+ri.y)/2.0;
         midmid.x = (mid.x+ri.x)/2.0;
         midmid.y = (mid.y+ri.y)/2.0;
         d1 = dis(t, mid)/r + dis(mid, d)/q;
         d2 = dis(t, midmid)/r + dis(midmid, d)/q;
         if(d1 > d2)
         le = mid;
         else ri = midmid;
     }while(dis(le, ri)>=eps);
     return d1;
 }

 double solve1()
 {
     double d1, d2;
     node le = a;
     node ri = b;
     node mid, midmid;
     do
     {
         mid.x = (le.x+ri.x)/2.0;
         mid.y = (le.y+ri.y)/2.0;
         midmid.x = (mid.x+ri.x)/2.0;
         midmid.y = (mid.y+ri.y)/2.0;
         d1 = dis(a, mid)/p + solve2(mid);
         d2 = dis(a, midmid)/p + solve2(midmid);
         if(d1 > d2)
         le = mid;
         else ri = midmid;
     }while(dis(le, ri)>=eps);
     return d1;
 }

 int main()
 {
     int t;
     scanf("%d", &t);
     while(t--)
     {
         scanf("%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y);
         scanf("%lf%lf%lf%lf", &c.x, &c.y, &d.x, &d.y);
         scanf("%lf%lf%lf", &p, &q, &r);
         printf("%.2lf\n", solve1());
     }
     return ;
 }