POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35043		Accepted: 10800

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<queue>

using namespace std;

const int maxn=100000;

int vis[maxn+10];
int n,k;

struct node{
    int x,c;
};

int BFS(){
    queue<node> q;
    while(!q.empty())
        q.pop();
    memset(vis,0,sizeof(vis));
    node cur,next;
    cur.x=n,cur.c=0;
    vis[cur.x]=1;
    q.push(cur);
    while(!q.empty()){
        cur=q.front();
        q.pop();
        for(int i=0;i<3;i++){
            if(i==0)
                next.x=cur.x-1;
            else if(i==1)
                next.x=cur.x+1;
            else
                next.x=cur.x*2;
            next.c=cur.c+1;
            if(next.x==k)
                return next.c;
            if(next.x>=0 && next.x<=maxn && !vis[next.x]){
                vis[next.x]=1;
                q.push(next);
            }
        }
    }
    return 0;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&n,&k)){
        if(n>=k){
            printf("%d\n",n-k);
            continue;
        }
        printf("%d\n",BFS());
    }
    return 0;
}