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.
1 #include<cstdio> 2 #include<queue> 3 #include<cstring> 4 using namespace std; 5 6 const int maxn=200050; 7 int N,K; 8 int vis[maxn]; 9 10 struct point 11 { 12 int pos; 13 int step; 14 }; 15 16 int bfs() 17 { 18 queue<point>q; 19 point init; 20 init.pos=N; 21 init.step=0; 22 q.push(init); 23 vis[N]=1; 24 while(!q.empty()) 25 { 26 point node=q.front(); 27 q.pop(); 28 if(node.pos==K) 29 return node.step; 30 point newnode1,newnode2,newnode3; 31 newnode1.pos=node.pos-1; 32 newnode2.pos=node.pos+1; 33 newnode3.pos=node.pos*2; 34 newnode1.step=newnode3.step=newnode2.step=node.step+1; 35 if(!vis[newnode1.pos]) 36 { 37 vis[newnode1.pos]=1; 38 q.push(newnode1); 39 } 40 if(!vis[newnode2.pos]) 41 { 42 vis[newnode2.pos]=1; 43 q.push(newnode2); 44 } 45 if(newnode3.pos<maxn&&newnode3.pos>=0) 46 if(!vis[newnode3.pos]) 47 { 48 vis[newnode3.pos]=1; 49 q.push(newnode3); 50 } 51 } 52 } 53 54 int main() 55 { 56 while(scanf("%d%d",&N,&K)!=EOF) 57 { 58 memset(vis,0,sizeof(vis)); 59 if(N>K) 60 printf("%d ",N-K); 61 else 62 { 63 int ans=bfs(); 64 printf("%d ",ans); 65 } 66 } 67 return 0; 68 }