| 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
Output
Sample Input
5 17
Sample Output
4
Hint
Source
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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()); } }} |
原文：http://www.cnblogs.com/ramanujan/p/3582822.html