Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 115041 | Accepted: 35968 |
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 minutës.
#include<iostream>
#include<stdlib.h>
#include<string>
#include<queue>
#include<stdio.h>
#include<string.h>
using namespace std;
int y;
int v[200005]={0};
int f[3][2]={2,0,1,-1,1,1};
struct node{
int x,t;
};
void bfs(int x,int t)
{
node s;
queue <node> q;
s.x=x;
s.t=t;
v[x]=1;
q.push(s);
while(!q.empty())
{
s=q.front();
q.pop();
if(s.x==y)
{
cout<<s.t<<endl;
break;
}
for(int i=0;i<3;i++)
{
int xx=s.x*f[i][0]+f[i][1];
if(xx>=0&&xx<=200001&&v[xx]==0)
{
v[xx]=1;
node ss;
ss.t=s.t+1;
ss.x=xx;
q.push(ss);
}
}
}
}
int main()
{
int x;
while(~scanf("%d%d",&x,&y))
{
memset(v,0,sizeof(v));
bfs(x,0);
}
}