搜索 G - 07

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.

          这道题的题意为给你两个位置，求第一个位置到第二个位置的最小步数（可以前走一步，后走一步，或向前走该位置数的两倍）。该题为BFS题，用队列储存每次三种走法的位置，然后从队头挨着搜，直到走到要求到的位置结束。

源代码如下：

#include<iostream>
 #include<string.h>
 #include<queue>
 using namespace std;
 int main()
 {
 	int n,k,m,v[100005];
 	while(cin>>n>>k)
 	{ memset(v,0,sizeof(v));
 	  m=0;
 	  if(n>=k)
	   cout<<n-k<<endl;
	  else { queue<int>q;
	  	     q.push(n);
	  	     v[n]=1;
	  	     q.push(m);
	  	     while(!q.empty())
	  	     {
	  	     	n=q.front();
	  	     	q.pop();
	  	     	m=q.front();
	  	     	q.pop();
	  	     	if(n==k) break;
	  	     	if(n+1<100005&&v[n+1]!=1)
	  	     	{
	  	     		q.push(n+1);
	  	     		v[n+1]=1;
	  	     		q.push(m+1);
			    }
			    
			    if(n-1>=0&&v[n-1]!=1)
	  	     	{
	  	     		q.push(n-1);
	  	     		v[n-1]=1;
	  	     		q.push(m+1);
			    }
			    
			    if(2*n<100005&&v[2*n]!=1)
	  	     	{
	  	     		q.push(2*n);
	  	     		v[2*n]=1;
	  	     		q.push(m+1);
			    }
			    
		     }
		     cout<<m<<endl;
	       }
	       
	 }
 }