Catch That Cow（BFS）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10166    Accepted Submission(s): 3179

Problem 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.

 #include <cstdio>
 #include <iostream>
 #include <queue>
 #include <cstring>
 using namespace std;
 int b,e;
 int Mintim;
 struct node
 {
     int pos,t;
 }k,tem;
 int vis[+];
 queue<node> s;
 void bfs()
 {
     while(!s.empty())
         s.pop();
     k.pos=b,k.t=;
     s.push(k);
     while(!s.empty())
     {
         k=s.front();
         s.pop();
         if(k.pos==e)
         {
             Mintim=k.t;
             return;
         }
         if(k.pos<||k.pos>||vis[k.pos])    continue;
         vis[k.pos]=;
         tem.t=k.t+;
         tem.pos=k.pos+;
         s.push(tem);
         tem.pos=k.pos-;
         s.push(tem);
         tem.pos=k.pos*;
         s.push(tem);
     }
 }
 int main()
 {
     int i,j;
     freopen("in.txt","r",stdin);
     while(scanf("%d%d",&b,&e)!=EOF)
     {
         memset(vis,,sizeof(vis));
         bfs();
         printf("%d\n",Mintim);
     }
     return ;
 }