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.

题目大意：

FJ要抓奶牛。

      开始输入N（FJ的位置）K（奶牛的位置）。

      FJ有三种移动方法：1、向前走一步，耗时一分钟。

                                       2、向后走一步，耗时一分钟。

                                       3、向前移动到当前位置的两倍N*2，耗时一分钟。

      问FJ抓到奶牛的最少时间。PS：奶牛是不会动的。

思路：1、如果FJ不在奶牛后面，那么他只有一步步往后移动到奶牛位置了，即N>=K时，输出N-K即可。

          2、否则bfs+队列查找

程序代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define maxn 100001
bool vis[maxn];
int step[maxn];
queue<int> q;
int bfs(int n,int k)
{
  int head,tail;
  q.push(n);
  step[n]=0;
  vis[n]=true;
  while(!q.empty())
  {
    head=q.front();
    q.pop();
    for(int i=0;i<3;i++)
    {
      if(i==0)
        tail=head-1;
      else if(i==1)
        tail=head+1;
      else
        tail=head*2;
      if(tail<0||tail>maxn)
        continue;
      if(!vis[tail])
      {
        q.push(tail);
        step[tail]=step[head]+1;
        vis[tail]=true;
      }
      if(tail==k)
        return step[tail];
    }
  }
}
int main()
{
  int n,k;
  while(cin>>n>>k)
  {
    memset(step,0,sizeof(step));
    memset(vis,false,sizeof(vis));
    while(!q.empty())
      q.pop();
    if(n>=k)
      cout<<n-k<<endl;
    else
      cout<<bfs(n,k)<<endl;
  }
  return 0;
}