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());
}
}
}