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Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 78708 | Accepted: 24827 |
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 <iostream>
#include <string.h>
#include <queue>
using namespace std;
struct node
{
int data,step;
};
int vis[200000];
void func(int n,int k)
{
struct node t,v;
queue<struct node>S;
t.step=0;
t.data=n;
vis[n]=1;
S.push(t);
while(!S.empty())
{
v=S.front();
S.pop();
if(v.data==k)
{
cout<<v.step<<endl;
return ;
}
if(v.data>=1&&!vis[v.data-1])
{
t.data=v.data-1;
t.step=v.step+1;
S.push(t);
vis[v.data-1]=1;
}
if(v.data<=k&&!vis[v.data+1])
{
t.data=v.data+1;
t.step=v.step+1;
S.push(t);
vis[v.data+1]=1;
}
if(v.data<=k&&!vis[v.data*2])
{
t.data=v.data*2;
t.step=v.step+1;
S.push(t);
vis[v.data*2]=1;
}
}
}
int main()
{
int n,k;
cin>>n>>k;
memset(vis,0,sizeof(vis));
func(n,k);
return 0;
}