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

 1 #include <iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<queue>
 5 #include<vector>
 6 #include<algorithm>
 7 using namespace std;
 8 struct node
 9 {
10     int cur,step;
11 }st;
12 queue<node>q;
13 int m,n,ans,visit[200000];
14 int dfs(int x)
15 {
16     int i;
17     while(!q.empty())
18     {
19         q.pop();
20     }
21     memset(visit,0,sizeof(visit));
22     node s,p;
23     q.push(st);
24     visit[st.cur]=1;
25     while(!q.empty())
26     {
27         p=q.front();
28         q.pop();
29         if(p.cur==n)
30             return p.step;
31         for(i=1;i<=3;i++)
32         {
33             s=p;
34             if(i==1)
35                s.cur-=1;
36             else if(i==2)
37                 s.cur+=1;
38             else
39                 s.cur*=2;
40             s.step++;
41             if(s.cur==n)
42                 return s.step;
43             if(s.cur>=0&&s.cur<=200000&&!visit[s.cur])
44             {
45                 visit[s.cur]=1;
46                 q.push(s);
47             }
48 
49         }
50     }
51 
52 }
53 int main()
54 {
55     while(scanf("%d%d",&m,&n)!=-1)
56     {
57         st.cur=m;
58         st.step=0;
59         ans=dfs(m);
60         printf("%d
",ans);
61     }
62     return 0;
63 }