题目链接：https://vjudge.net/problem/POJ-3278
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.

题目大意： 花花经过长时间的研究，终于研发出了能够跃迁的宇宙飞船。现在，他想要前往致远星。假设地球和致远星都在一个坐标轴上，其中地球位于坐标n，而致远星位于坐标k。而花花的飞船支持以下两种运动方式：

    飞行：在一个时间单位中，能够从坐标x移动到x-1或x+1；
    跃迁：在一个时间单位中，能够直接从x跃迁到2x。

    现在，花花想知道，他需要多长时间才能到达致远星？

解题思路：利用BFS求出n到k的最短距离，关键在于对搜索过程进行剪枝，省去重复的工作。

Cpp Code

#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 1e5 + 5;
int n, k;
bool vis[maxn];//标记数组
int ans[maxn];
queue<int> q;
int bfs(int x,int k){
    int cur, next;
    ans[x] = 0;
    vis[x] = 1;//访问后标记为1，防止重复访问
    q.push(x);
    while (!q.empty())
    {
        cur = q.front();
        q.pop();
        for (int i = 0; i < 3;i++){//三次转换
            if(i==0){
                next = cur - 1;
            }
            else if(i==1){
                next = cur + 1;
            }
            else{
               next = cur*2; 
            }
            //符合条件才压入队列
            if(next>=0&&next<=100000&&vis[next]==0){
                vis[next] = 1;
                q.push(next);
                ans[next] = ans[cur] + 1;
            }
            if(next==k){
                return ans[next];
            }
        }
        /* code */
    }
    
}
int main(){
    cin >> n >> k;
    fill(vis, vis + maxn, 0);
    while (!q.empty())
    {
        q.pop();
    }
    if(n>=k){
        cout << n - k << endl;
    }
    else{
        cout << bfs(n,k) << endl;
    }
    return 0;
}

Java Code

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;


public class Main {
	static class node{
		int x;//数的值
		int num;//次数
		node(){}
		node(int a,int b){
			x=a;
			num=b;
		}
	}
	static boolean[] vis=new boolean[(int)1e5+5];//标记数组
	
	static boolean ok(int x) {//判断函数
		if(x>=0&&x<=(int)1e5&&vis[x]==false) {
			return true;
		}
		return false;
	}
	public static int BFS(int n,int k) {
		
		Queue<node> q=new LinkedList<node>();
		vis[n]=true;//访问后标记
		q.offer(new node(n,0));
		node op;
		while(!q.isEmpty()) {
			op=q.poll();
			if(op.x==k) {
				return op.num;
			}
			int x,num=op.num+1;
			for(int i=0;i<3;i++) {//三次转换
				if(i==0) {
					x=op.x-1;
				}
				else if(i==1) {
					x=op.x+1;
				}
				else {
					x=op.x*2;
				}
				if(ok(x)) {
					vis[x]=true;//访问后标记
					q.offer(new node(x,num));
				}
			}
			
		}
		return -1;
	}
	
	
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n,k;
		n=sc.nextInt();
		k=sc.nextInt();
		System.out.print(BFS(n,k));
		// TODO Auto-generated method stub

	}

}