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 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.

通俗讲就是  n可以+1可以-1可以*2  让n=k  最少要几步  

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int vis[1000001];
struct node
{
	int wei;
	int time;
};
int main()
{
	int n,k;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		if(n==k)
		{
			printf("0\n");
			continue;
		}
		memset(vis,0,sizeof(vis));
	   struct node start;
	    	start.wei =n;
	    	start.time =0;
	    	vis[n]=1;
	    	queue<node>q;
	    	q.push(start);
			while(!q.empty() )
			{
				struct node tmp=q.front();
				q.pop(); 
				if(tmp.wei-1==k||tmp.wei +1==k||tmp.wei *2==k)
				{
					printf("%d\n",tmp.time+1 );
					break;
				}
				tmp.time++;
					struct node tmp2;
				if(tmp.wei +1<k)
				{
			     	tmp2=tmp;
				    tmp2.wei ++;
				   if(vis[tmp2.wei]==0)
				   {
				   		vis[tmp2.wei]=1;
					q.push(tmp2) ;
				   }
				}
			   if(tmp.wei-1>0&&tmp.wei-1!=k)
				{
			     	tmp2=tmp;
				    tmp2.wei--;
				   if(vis[tmp2.wei ]==0)
				   {
				   		vis[tmp2.wei]=1;
					   q.push(tmp2) ;
				   }
				}
					if(tmp.wei<k)
				{
			     	tmp2=tmp;
				    tmp2.wei=2*tmp.wei
					;
				   if(vis[tmp2.wei]==0)
				   {
				   		vis[tmp2.wei]=1;
					q.push(tmp2) ;
				   }
				}
			}
	}
	return 0;
}