Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10166 Accepted Submission(s): 3179
Problem 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?
* 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 <iostream> 3 #include <queue> 4 #include <cstring> 5 using namespace std; 6 int b,e; 7 int Mintim; 8 struct node 9 { 10 int pos,t; 11 }k,tem; 12 int vis[100000+10]; 13 queue<node> s; 14 void bfs() 15 { 16 while(!s.empty()) 17 s.pop(); 18 k.pos=b,k.t=0; 19 s.push(k); 20 while(!s.empty()) 21 { 22 k=s.front(); 23 s.pop(); 24 if(k.pos==e) 25 { 26 Mintim=k.t; 27 return; 28 } 29 if(k.pos<0||k.pos>100000||vis[k.pos]) continue; 30 vis[k.pos]=1; 31 tem.t=k.t+1; 32 tem.pos=k.pos+1; 33 s.push(tem); 34 tem.pos=k.pos-1; 35 s.push(tem); 36 tem.pos=k.pos*2; 37 s.push(tem); 38 } 39 } 40 int main() 41 { 42 int i,j; 43 freopen("in.txt","r",stdin); 44 while(scanf("%d%d",&b,&e)!=EOF) 45 { 46 memset(vis,0,sizeof(vis)); 47 bfs(); 48 printf("%d ",Mintim); 49 } 50 return 0; 51 }