思路：bfs裸题。三个选择：向左一个单位，向右一个单位，向右到2*x

//注意，需要特判n是否大于k，大于k时只能向左，输出n-k。第一次提交没注意，结果RE了，，

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 137789		Accepted: 42551

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.

Source

USACO 2007 Open Silver

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
class Point{
	public int x;
	public int count;
	public Point(int x,int y){
		this.x=x;
		this.count=y;
	}
	public Point(){}
}
public class Main{

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in=new Scanner(System.in);
		int n=in.nextInt();
		int k=in.nextInt();
		boolean vis[]=new boolean[k*2];
		if(n>=k)System.out.println(n-k);
		else 
			System.out.println(bfs(n,k,vis));
	}

	private static int bfs(int xx, int k,boolean vis[]) {
		Queue<Point> q=new LinkedList<Point>();
		Point p=new Point(); 
		p.x=xx;
		p.count=0;
		vis[p.x]=true;
		q.add(p);
		while(!q.isEmpty()){			
			Point n=q.remove();
			if(n.x>=2*k||n.x<0)continue;

			if(n.x==k){
				return n.count;
			}else{
				
				for(int i=0;i<3;i++)
				{
					Point n1=new Point();
					if(i==0){
						n1.x=n.x+1;
					}
					else if(i==1)
						n1.x=n.x-1;
					else
						n1.x=2*n.x;
					if((!(n1.x>=2*k||n1.x<0))&&vis[n1.x]==false)
					{
						vis[n1.x]=true;
						n1.count=n.count+1;
						q.add(n1);

					}
				}
			}
			
		}
		return 0;
	}

}