B - Catch That Cow

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.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
int vis[101000];
struct node
{
    int x,time;
}tail,head;
int main ()
{
    int n,k;
    cin>>n>>k;
    memset(vis,0,sizeof(vis));
    queue <node> s;
    head.x=n;
    head.time=0;
    vis[n]=1;
    s.push(head);
    while(!s.empty())
    {
        head=s.front();
        s.pop();
        int i;
        if(head.x==k)
        {
            printf("%d\n",head.time);
            break;
        }
        for(i=0;i<3;i++)
        {
            if(i==0)
            {
                tail.x=head.x+1;
                tail.time=head.time+1;
            }
            else if(i==1)
            {
                tail.x=head.x-1;
                tail.time=head.time+1;
            }
            else if(i==2)
            {
                tail.x=head.x*2;
                tail.time=head.time+1;
            }
            if(tail.x<=101000&&tail.x>=0&&vis[tail.x]==0)
            {
                vis[tail.x]=1;
                s.push(tail);
            }
        }
    }
}