bfs hdu 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14553 Accepted Submission(s): 4422

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?

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

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<iostream>
 2 #include<queue>
 3 #include<string.h>
 4 
 5 using namespace std;
 6 
 7 int k;
 8 int vis[100005];
 9 struct node
10 {
11     int data;
12     int step;
13 };
14 node n;
15 int bfs()
16 {
17     queue<node>que;
18     que.push(n);
19     while(!que.empty())
20     {
21         node p = que.front();
22         que.pop();
23         vis[p.data]=1;
24         if(p.data==k)
25         {
26             return p.step;
27         }
28         node q = p;
29         q.step++;
30         
31         q.data = p.data*2;
32         if(q.data>=0 && q.data<=100000 && !vis[q.data])
33             que.push(q);
34         q.data=p.data-1;
35         if(q.data>=0 && q.data<=100000 && !vis[q.data])
36             que.push(q);
37         q.data=p.data+1;
38         if(q.data>=0 && q.data<=100000 && !vis[q.data])
39             que.push(q);
40         
41     }
42     return -1;
43 }
44 int main()
45 {
46     while(cin>>n.data>>k)
47     {
48         memset(vis,0,sizeof(vis));
49         cout<<bfs()<<endl;
50     }
51     
52     
53     return 0;
54 }