poj3278Catch That Cow

Catch That Cow

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

USACO 2007 Open Silver

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

int que[200001];
int start=0,endd=0;
int time[200001]={0};
int vis[200001]={0};
int n,k;

int bfs(int n,int k){
    memset(que,0,sizeof(que));
    memset(time,0,sizeof(time));
    memset(vis,0,sizeof(vis));
    start=0;
    endd=0;
    que[endd++]=n;
    vis[n]=1;
    while(start<endd){
        int t=que[start];
        start++;
        for(int i=0;i<3;i++){
            int tt=t;
            if(i==0){
                tt+=1;
            }else if(i==1){
                tt-=1;
            }else if(i==2){
                tt*=2;
            }
            if(tt>100000||tt<0){
                continue;
            }
            if(!vis[tt]){
                vis[tt]=1;
                que[endd]=tt;
                time[tt]=time[t]+1;
                if(tt==k){
                    return time[tt];
                }
                endd++;
            }
        }
    }
}

int main()
{
    int ans;
    while(~scanf("%d %d",&n,&k)){
        if(n<k){
            ans=bfs(n,k);
        }else{
            ans=n-k;
        }
        printf("%d\n",ans);
    }
    return 0;
}