【搜索】POJ3278：Catch That Cow

Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN(0 ≤N≤ 100,000) on a number line and the cow is at a pointK(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 pointXto the pointsX- 1 orX+ 1 in a single minute
* Teleporting: FJ can move from any pointXto the point 2 ×Xin 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:NandK

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.

1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <queue>
 6 using namespace std;
 7 const int maxn = 100010;
 8 bool vis[maxn];
 9 int step[maxn];
10 
11 bool bound(int num)
12 {
13     if(num<0||num>100000)
14         return true;
15     return false;
16 }
17 int bfs(int sta,int endd)
18 {
19     queue<int> q;
20     int now,nextt;
21     q.push(sta);
22     vis[sta]=true;
23     while(!q.empty())
24     {
25         now=q.front();
26         q.pop();
27         for(int i=0;i<3;i++)
28         {
29             if(i==0)
30                 nextt=now+1;
31             else if(i==1)
32                 nextt=now-1;
33             else
34                 nextt=now*2;
35             if(bound(nextt))
36                 continue;
37             if(!vis[nextt])
38             {
39                 step[nextt]=step[now]+1;
40                 if(nextt==endd)
41                     return step[nextt];
42                 vis[nextt]=true;
43                 q.push(nextt);
44             }
45         }
46     }
47 }
48 int main()
49 {
50     int sta,endd;
51     while(scanf("%d%d",&sta,&endd)!=EOF)
52     {
53         memset(vis,false,sizeof(vis));
54         if(sta>=endd)
55             printf("%d\n",sta-endd);
56         else
57         {
58             int ans=bfs(sta,endd);
59             printf("%d\n",ans);
60         }
61     }
62     return 0;
63 }