poj 3278 Catch That Cow(bfs+队列)

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.

　题意：FJ要抓奶牛，输入n（FJ的位置）,k（奶牛的位置）,求FJ抓到奶牛所需最少时间。

　FJ有三种走法：

　　　　1：向前移动一步，花费一分钟

　　　　2：向后退后一步，花费一分钟

　　　  3：向前移动当前位置的2倍，花费一分钟

 #include <iostream>
 #include <cstring>
 #include <queue>
 using namespace std;
 const int MAX=;
 bool vis[MAX];
 int step[MAX];
 queue <int >q;
 int bfs(int n,int k)
 {
     int head,next;
     q.push(n);
     step[n]=;
     vis[n]=true;
     while(!q.empty())
     {
         head=q.front();  //取队首
         q.pop();
         for(int i=; i<; i++)
         {
             if(i==)
                 next=head-;
             if(i==)
                 next=head+;
             if(i==)
                 next=head*;
             if(next<||next>=MAX)  //排除出界情况
                 continue;
             if(!vis[next])
             {
                 q.push(next);  //入队
                 step[next]=step[head]+;//步数加一
                 vis[next]=true;  //标记访问
             }
             if(next==k)
                 return step[next];
         }

     }
 }
 int main()
 {
     int n,k;
     while(cin>>n>>k)
     {
         memset(vis,false,sizeof(vis));
         memset(step,,sizeof(step));
         while(!q.empty())
             q.pop();
         if(n>=k)
             cout<<n-k<<endl;
         else
             cout<<bfs(n,k)<<endl;
     }
     return ;
 }