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
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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;
}