Catch That Cow（bfs)

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
此题需要注意的是queue应该定义全局，这一点上我wa了好几发

 #include <iostream>
 #include<stdio.h>
 #include<string.h>
 #include<queue>
 #include<vector>
 #include<algorithm>
 using namespace std;
 struct node
 {
     int cur,step;
 }st;
 queue<node>q;
 int m,n,ans,visit[];
 int dfs(int x)
 {
     int i;
     while(!q.empty())
     {
         q.pop();
     }
     memset(visit,,sizeof(visit));
     node s,p;
     q.push(st);
     visit[st.cur]=;
     while(!q.empty())
     {
         p=q.front();
         q.pop();
         if(p.cur==n)
             return p.step;
         for(i=;i<=;i++)
         {
             s=p;
             if(i==)
                s.cur-=;
             else if(i==)
                 s.cur+=;
             else
                 s.cur*=;
             s.step++;
             if(s.cur==n)
                 return s.step;
             if(s.cur>=&&s.cur<=&&!visit[s.cur])
             {
                 visit[s.cur]=;
                 q.push(s);
             }

         }
     }

 }
 int main()
 {
     while(scanf("%d%d",&m,&n)!=-)
     {
         st.cur=m;
         st.step=;
         ans=dfs(m);
         printf("%d\n",ans);
     }
     return ;
 }