（HDU - 3400）Line belt

最新推荐文章于 2018-03-14 18:58:07 发布

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该问题描述了一个在二维平面上从点A到点D的路径规划，点A和点B、点C和点D分别是两个线形传送带的端点。物体在传送带AB上的速度为P，CD上的速度为Q，其他区域速度为R。目标是找到从A到D的最短时间。解决方案涉及到在传送带之间找到转折点以最小化总旅行时间。

（HDU - 3400）Line belt

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4640 Accepted Submission(s): 1796

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

题目大意：二维平面上有两条传送ab，cd，物体在传送带ab上的速度是p，在传送带cd上的速度是q，在其他地方的速度是r，问物体从a到d所需的时间最小是多少。

思路：最小的时间必然是在ab上走一段，cd上走一段，ab，cd之间走一段，设ab上的转折点为m，cd上的转折点为n，则时间 $t=am/p+mn/r+nd/q$ 。假设m点确定，那么在cd之间三分即可求出点n使 $mn/r+nd/q$ 最小，再在ab之间三分求整体时间最小，即三分嵌套。

这里写图片描述

#include<cstdio>
#include<cmath>
using namespace std;

const double eps=1e-8;
double p,q,r;

struct node
{
    double x,y;
};

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

node getmid(node a,node b)
{
    node c;
    c.x=(a.x+b.x)/2.0;
    c.y=(a.y+b.y)/2.0;
    return c;
}

double ts2(node m,node c,node d)//确定m在c,d之间ternary_search(三分)
{
    node lo=c,hi=d,mid,midmid;
    double t1=0,t2=1;//给t1,t2赋初值是为了防止刚开始t1，t2就很接近而不进行while循环 
    while(fabs(t1-t2)>eps)
    {
        mid=getmid(lo,hi);
        midmid=getmid(mid,hi);
        t1=dis(m,mid)/r+dis(mid,d)/q;
        t2=dis(m,midmid)/r+dis(midmid,d)/q;
        if(t1<t2) hi=midmid;
        else lo=mid; 
    } 
    return t1;
} 

double ts1(node a,node b,node c,node d)//再a,b之间整体三分 
{
    node lo=a,hi=b,mid,midmid;
    double t1=0,t2=1;
    while(fabs(t1-t2)>eps)
    {
        mid=getmid(lo,hi);
        midmid=getmid(mid,hi);
        t1=dis(a,mid)/p+ts2(mid,c,d);
        t2=dis(a,midmid)/p+ts2(midmid,c,d);
        if(t1<t2) hi=midmid;
        else lo=mid;
    }
    return t1;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        node a,b,c,d;
        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("%.2f\n",ts1(a,b,c,d));
    }
    return 0;
}