题目：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.

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
bool vis[1000030];//尽量开大 否则数组越界	runtime error
	int n,k;
queue<int>q;
int count[1000030];
void bfs(int n,int k)
{
   
	q.push(n);
	vis[n]=1;
	count[n]=0;
	while(!q.empty())
	{
   
		int x=q.front();
		q.pop();
		if(k==x)
		{
   
			printf("%d\n",count[x]);
			break;
		}
		if(x-1>=0&&!vis[x-1])
		{
   
			vis[x-1]=1;
			q.push(x-1);
			count[x-1]=count[x]+1;	
		}
		if(x<=k&&!vis[x+1])
		{
   	
			vis[x+1]=1;
			q.push(x+1);
			count[x+1]=count[x]+1;
		}
		if(x<=k&&!vis[2*x])
		{
   
			q.push(2*x);
			vis[2*x]=1;
			count[2*x]=count[x]+1;
		}
	}		
	
}
int main()
{
   
	memset(vis,0,sizeof(vis));
	scanf("%d%d",&n,&k);
	bfs(n,k); 
}