poj 3278 bfs
Catch That Cow
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 41703 | Accepted: 13005 |
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
练习怎用队列
import java.awt.Point; import java.util.LinkedList; import java.util.Queue; import java.util.Scanner; class Bfs{ private static final int SIZE=200000; private int s, t; private boolean[] vis; private Queue<Point> q; public Bfs(int s, int t){ this.s=s; this.t=t; q = new LinkedList<Point>(); vis = new boolean[SIZE]; for(int i=0; i<SIZE; i++) vis[i]=false; } public int run(){ q.add(new Point(s,0)); while(!q.isEmpty()){ Point cur = q.poll(); if(cur.x == t) return cur.y; int y = cur.y+1; if(cur.x>0 && !vis[cur.x-1]){ q.add(new Point(cur.x-1,y)); vis[cur.x-1]=true; } if((cur.x<<1)<SIZE && !vis[cur.x<<1]){ q.add(new Point(cur.x<<1,y)); vis[cur.x<<1]=true; } if(cur.x+1<SIZE && !vis[cur.x+1]){ q.add(new Point(cur.x+1, y)); vis[cur.x+1]=true; } } return SIZE>>1; } } public class Main { static final int INF = 200000; /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner s = new Scanner(System.in); while(s.hasNext()){ int n = s.nextInt(), k=s.nextInt(); Bfs bfs = new Bfs(n,k); System.out.println(bfs.run()); } } }