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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 |
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 }