Catch That Cow（BFS）

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 62027		Accepted: 19403

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

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<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int vis[200005];
int s,e;
struct node
{
    int cur;
    int step;
};
int check(node a)
{
    if(a.cur>200005||a.cur<0||vis[a.cur])
    {
        return 0;
    }
    return 1;
}
int bfs(int p)
{
    queue<node>Q;
    node rencent,next;
    rencent.cur=p;
    rencent.step=0;
    vis[rencent.cur]=1;
    Q.push(rencent);
    while(!Q.empty())
    {
        rencent=Q.front();
        Q.pop();
        if(rencent.cur==e)
        {
            return rencent.step;
        }
        next.cur=rencent.cur+1;
        if(check(next))
        {
            vis[next.cur]=1;
            next.step=rencent.step+1;
            Q.push(next);
        }
        next.cur=rencent.cur-1;
        if(check(next))
        {
            vis[next.cur]=1;
            next.step=rencent.step+1;
            Q.push(next);
        }
        next.cur=rencent.cur*2;
        if(check(next))
        {
            vis[next.cur]=1;
            next.step=rencent.step+1;
            Q.push(next);
        }
    }
}
int main()
{
    while(cin>>s>>e)
    {
        memset(vis,0,sizeof(vis));
        cout<<bfs(s)<<endl;
    }
}