Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 88361 | Accepted: 27679 |
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.
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue> using namespace std; int que[];
int start=,endd=;
int time[]={};
int vis[]={};
int n,k; int bfs(int n,int k){
memset(que,,sizeof(que));
memset(time,,sizeof(time));
memset(vis,,sizeof(vis));
start=;
endd=;
que[endd++]=n;
vis[n]=;
while(start<endd){
int t=que[start];
start++;
for(int i=;i<;i++){
int tt=t;
if(i==){
tt+=;
}else if(i==){
tt-=;
}else if(i==){
tt*=;
}
if(tt>||tt<){
continue;
}
if(!vis[tt]){
vis[tt]=;
que[endd]=tt;
time[tt]=time[t]+;
if(tt==k){
return time[tt];
}
endd++;
}
}
}
} int main()
{
int ans;
while(~scanf("%d %d",&n,&k)){
if(n<k){
ans=bfs(n,k);
}else{
ans=n-k;
}
printf("%d\n",ans);
}
return ;
}