HDU 2717 Catch That Cow(bfs)

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10575    Accepted Submission(s): 3303     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 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 USACO 2007 Open Silver

 

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <queue>
 5 using namespace std;
 6 long long l,k;
 7 const int maxn=100000;
 8 bool v[100010];
 9 struct  node
10 {
11     long long x;
12     int step;
13 };
14 bool check(long long a)
15 {
16     if(a<0||a>maxn||v[a]) return false;
17     return true;
18 }
19 int bfs()
20 {
21     node n,m;
22     n.x=l;
23     n.step=0;
24     queue<node> q;
25     q.push(n);
26     v[n.x]=1;
27     while(!q.empty()){
28         n=q.front();
29         q.pop();
30         if(n.x==k) return n.step;
31         //-1
32         m.x=n.x-1;
33         if(check(m.x)){
34             m.step=n.step+1;
35             v[m.x]=1;
36             q.push(m);
37         }
38         //+1
39         m.x=n.x+1;
40         if(check(m.x)){
41             m.step=n.step+1;
42             v[m.x]=1;
43             q.push(m);
44         }
45         //*2
46         m.x=n.x*2;
47         if(check(m.x)){
48             m.step=n.step+1;
49             v[m.x]=1;
50             q.push(m);
51         }
52     }
53     return -1;
54 }
55 int main()
56 {
57     while(cin>>l>>k){
58         memset(v,0,sizeof(v));
59         cout<<bfs()<<endl;
60     }
61     return 0;
62 }