Hurry Up（三分）

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本文介绍了一道算法题目，探讨了如何计算从任意位置到目的地的最短时间，包括直接跑步和乘坐出租车两种方式。提供了两种解决方案，一种是通过三分法进行优化，另一种是通过枚举法求解。

http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1175

(2013湘潭邀请赛题目)

Hurry Up

Accepted : 62

Submit : 251

Time Limit : 1000 MS

Memory Limit : 65536 KB

Problem Description

GG is some what afraid of his MM. Once his MM asks, he will always try his best to rush to their home.
Obvious, he can run home in straight line directly. Alternatively, he can run to the main road and call the taxi.
You can assume there is only one main road on the x-axis, with unlimited length.
Given the initial location of GG and his destination, please help to ask the minimize time to get home.
GG will always run at the fixed speed of vr, and the taxi can move at the fixed speed of vt
You can also assume that, only GG reach the main road, he can catch the taxi immediately. And the taxi will go towards home ( not necessay along the road ).
Bisides, GG can run arbitrary length, and pay arbitrarily for the taxi.

P.S. MM: abbr. Ma Ma

Input

Multiple test cases. First line, an integer T ( 1 ≤ T ≤ 2000 ), indicating the number of test cases.
For each test cases, there are 6 integers x1, y1, x2, y2, vr, vt in a line.
( -1000 <= x1, y1, x2, y2 <= 1000, 1 <= vr < vt <= 1000 )
(x1, y1) : the initial location of GG
(x2, y2) : the destination location of GG
vr: GG's run speed
vt: taxi's speed

Ouput

For each test case, output a real number with 2 digits after the arithmetic point. It is the shorest time for GG to reach home.

Sample Input

1 1 2 2 1 2

1 1 2 2 1 7

Sample Output

1.41

1.32

Source

XTU OnlineJudge

[ Submit Solution ]

//正解一

memory      time             length

1064 KB  31 MS      1123 B

#include<string.h>

#include<stdio.h>

#include<math.h>

#include<algorithm>

#include <iostream>

using namespace std;

const double ep=1e-10;

double vr,vt;

double xt,x2;

double y2,yt;

double lenth(double a1,double b1,double a2,double b2)

{

    return sqrt((a1-a2)*(a1-a2)+(b1-b2)*(b1-b2));

}

double caltime(double x)

{

    return (lenth(xt,yt,x,0)/vr+lenth(x2,y2,x,0)/vt);

}

int main()

{

    int T,i;

    double t1,t2,r,l;

    while(scanf("%d",&T)!=EOF)

    {

        for(i=0;i<T;i++)

        {

            scanf("%lf%lf%lf%lf",&xt,&yt,&x2,&y2);

            scanf("%lf%lf",&vr,&vt);

            t1=lenth(xt,yt,x2,y2)/vr;

            r=xt;

            l=x2;

              for(int i=0;i<100;i++)//这么多是参考某位大神的

           {  double m1=l + (r-l)/3;

             double m2=r - (r-l)/3;

             if(caltime(m1)<=caltime(m2))  r=m2;

             else  l=m1;

           }

             t2=caltime(l);

              t1=t1<t2? t1:t2;

             printf("%.2lf\n",t1);

        }

    }

    return 0;

}

正解二：/

枚举法

memory      time             length
1256 KB     593 MS      1176 B


//枚举法，枚举到0.1

#include<string.h>

#include<stdio.h>

#include<math.h>

#include<algorithm>

#include <iostream>

using namespace std;

const double ep=1e-10;

double vr,vt;

double xt,x2;

double y2,yt;

double lenth(double a1,double b1,double a2,double b2)

{

    return sqrt((a1-a2)*(a1-a2)+(b1-b2)*(b1-b2));

}

double caltime(int x)

{    (double)x;

    return (lenth(xt,yt,x,0)/vr+lenth(x2,y2,x,0)/vt);

}

int main()

{

    int T,i,j;

    double t1,t2,min,max,t;

    while(scanf("%d",&T)!=EOF)

    {

        for(i=0;i<T;i++)

        {

            scanf("%lf%lf%lf%lf",&xt,&yt,&x2,&y2);

            xt=xt*10.0;

            yt=yt*10.0;

            x2=x2*10.0;

            y2=y2*10.0;

            scanf("%lf%lf",&vr,&vt);

            vr=vr*10.0;

            vt=vt*10.0;


            t1=lenth(xt,yt,x2,y2)/vr;

            max=xt>x2? xt:x2;

            min=xt<x2? xt:x2;

            t2=1000000.0;

            for(j=(int)min;j<=(int)max;j++)

            {

              t=caltime(j);

              if(t2>t)

              t2=t;



            }


              t1=t1<t2? t1:t2;

             printf("%.2lf\n",t1);

        }


    }

    return 0;

}

错误：

//三分法

#include<string.h>

#include<stdio.h>

#include<math.h>

#include<algorithm>

#include <iostream>

using namespace std;

const double ep=1e-10;

double vr,vt;

double xt,x2;

double y2,yt;

double lenth(double a1,double b1,double a2,double b2)

{

    return sqrt((a1-a2)*(a1-a2)+(b1-b2)*(b1-b2));

}

double caltime(double x)

{

    return (lenth(xt,yt,x,0)/vr+lenth(x2,y2,x,0)/vt);

}

double solve(double l,double r)//三分法

{

    double m;

    double mm;

    while(l+ep<r)

    {

        m=(l+r)/2.0;

        mm=(m+r)/2.0;

        if(caltime(m)<caltime(mm))//取最小值

         r=mm;

         else

         l=m;

    }

    return l;

}

int main()

{

    int T,i;

    double t1,t2;

    while(scanf("%d",&T)!=EOF)

    {

        for(i=0;i<T;i++)

        {

            scanf("%lf%lf%lf%lf",&xt,&yt,&x2,&y2);

            scanf("%lf%lf",&vr,&vt);

            t1=lenth(xt,yt,x2,y2)/vr;

             t2=caltime(solve(xt,x2));

              t1=t1<t2? t1:t2;

             printf("%.2lf\n",t1);

        }


    }

    return 0;

}