POJ 3278, Catch That Cow

Time Limit: 2000MS  Memory Limit: 65536K
Total Submissions: 13066  Accepted: 3968

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

 

Input
Line 1: Two space-separated integers: N and K

 

Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

 

Sample Input
5 17

 

Sample Output
4

Hint
The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

 

Source
USACO 2007 Open Silver

---

//  POJ3278.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < queue >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
     int  N, K;
    cin  >>  N  >>  K;

     const   int  MAXSIZE  =   200002 ;
     int  line[MAXSIZE];
    memset(line,  - 1 ,  sizeof (line));

    queue < int >  q;
    q.push(N);
    line[N]  =   0 ;

     while ( ! q.empty())
    {
         int  stp  =  q.front();
        q.pop();

         int  next  =  stp  -   1 ;
         if  (next  >=   0   &&  line[next]  ==   - 1 )
        {
            line[next]  =  line[stp]  +   1 ;
            q.push(next);
             if  (next == K)
                 break ;
        }

        next  =  stp  +   1 ;
         if  (next  <  MAXSIZE  &&  line[next]  ==   - 1 )
        {
            line[next]  =  line[stp]  +   1 ;
            q.push(next);
             if  (next == K)
                 break ;
        }

        next  =  stp  <<   1 ;
         if  (next  <  MAXSIZE  &&  line[next]  ==   - 1 )
        {
            line[next]  =  line[stp]  +   1 ;
            q.push(next);
             if  (next == K)
                 break ;
        }
    }

    cout  <<  line[K]  <<  endl;
     return   0 ;
}