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.
#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=100001;
bool a[maxn];
int step[maxn];
queue <int> q;
int bfs(int n,int k)
{
int head,next;
q.push(n);
step[n]=0;
a[n]=1;
while(!q.empty())
{
head=q.front();
q.pop();
for(int i=0; i<3; i++)
{
if(i==0) next=head-1;
else if(i==1) next=head+1;
else next=head*2;
if(next<0||next>=maxn) continue;
while(!a[next])
{
a[next]=1;
step[next]=step[head]+1;
q.push(next);
}
if(next==k) return step[next];
}
}
}
int main()
{
int n,k;
while(cin>>n>>k)
{
memset(a,0,sizeof(a));
memset(step,0,sizeof(step));
if(n>k) printf("%d\n",n-k);
else printf("%d\n",bfs(n,k));
}
return 0;
}
