Hdu2717 Catch That Cow(BFS) ---Java版

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13472 Accepted Submission(s): 4143

Problem 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.

题意：将给定的数字n通过 +1、-1、*2 变为k，输出最小变换次数。

思路：BFS即可，Java的BFS不太熟悉，开始打算用List记录访问过的数字，可是超时了，看来还是要用数组标记。队列要自己写，结构体换成类，入队列可以用构造传参。

import java.util.ArrayList;import java.util.List;import java.util.Scanner;public class Main {public static void main(String[] args) {       Scanner sc = new Scanner(System.in);       while(sc.hasNext()){       int n = sc.nextInt();       int k = sc.nextInt();       int sum=0;       sum = bfs(n,k);       System.out.println(sum);       }}public static int bfs(int n, int k) {Queue que = new Queue();       Node node = new Node();       Boolean vis[] = new Boolean[200001];       for(int i=0;i<vis.length;i++){       vis[i]=false;       }       int sum=0;       node.step=0;       node.num=n;       que.push(node);       vis[n]=true;       while(que.size()>0){       Node temp = (Node) que.pop();               if(temp.num==k){        sum=temp.step;        break;        }              int t = temp.num;      if(t-1>=0 && !vis[t-1]){      Node te = new Node();      te.num=t-1;      te.step=temp.step+1;      que.push(te);      vis[te.num]=true;      }            if(t+1<200000 && !vis[t+1]){      Node te = new Node();      te.num=t+1;      te.step=temp.step+1;      que.push(te);      vis[te.num]=true;      }            if(2*t<200000&&!vis[2*t]){      Node te = new Node();      te.num=t*2;      te.step=temp.step+1;      que.push(te);      vis[te.num]=true;      }       }return sum;}}class Queue{List list = new ArrayList();public void push(Object obj){list.add(obj);}public Object pop(){if(list.size()>0){return list.remove(0);}else{return null;}}public int size(){return list.size();}}class Node{int num;int step;}