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;
    }

}