hdu 3400Line belt(三分法)

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2517    Accepted Submission(s): 961

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：

给出两条线段AB和CD，在AB上运动有一个速度，在CD上运动有一个速度，两条线段之外有一个速度。现在从A点出发，到达D点。求所需最短时间。

解题思路：

三分法，找出AB上运动到D点的最短时间的点，再找出在CD上的点。那么路线A到D之间经过这两个点的时间就是最短的。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif
#define eps 1e-8//注意精度问题
struct point
{
    double x,y;
};

double p,q,r;

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

double fund(point a,point c,point d)
{
    point mid;
    point midmid;
    point left=c;
    point right=d;
    double mid_value,midmid_value;
    int flag=0;
    while(dis(left,right)>=eps)
    {
        mid.x=(left.x+right.x)/2.0;
        mid.y=(left.y+right.y)/2.0;
        midmid.x=(mid.x+right.x)/2.0;
        midmid.y=(mid.y+right.y)/2.0;
        mid_value=dis(a,mid)/r+dis(mid,d)/q;
        midmid_value=dis(a,midmid)/r+dis(midmid,d)/q;
        if(mid_value<=midmid_value)
            right=midmid;
        else left=mid;
        flag=1;
    }
    if(flag)
        return mid_value;
    else
        return dis(a,d)/r;
}

double ans(point a,point b,point c,point d)
{
    point mid,midmid;
    point left=a;
    point right=b;
    double mid_value,midmid_value;
    int flag=0;
    while(dis(left,right)>=eps)
    {
        mid.x=(left.x+right.x)/2.0;
        mid.y=(left.y+right.y)/2.0;
        midmid.x=(mid.x+right.x)/2.0;
        midmid.y=(mid.y+right.y)/2.0;
        mid_value=dis(mid,a)/p+fund(mid,c,d);
        midmid_value=dis(midmid,a)/p+fund(midmid,c,d);
        if(mid_value<=midmid_value)
            right=midmid;
        else left=mid;
        flag=1;
    }
    if(flag)
        return mid_value;
    else
        return fund(a,c,d);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        point 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("%.2lf\n",ans(a,b,c,d));
    }
    return 0;
}