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Catch That Cow

2024-09-13 13:14:00   218人阅读

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 <iostream>
 2 #include <queue>
 3 using namespace std;
 4 
 5 struct node{
 6     int loc, step;//loc 当前位置， step 次数
 7 }pos,q;
 8 
 9 queue<node> que;
10 int vis[200005];
11 
12 void Push(int loc, int step){
13     q.loc = loc;
14     q.step = step + 1;
15     vis[loc] = 1;
16     que.push(q);
17 }
18 
19 int bfs(int n, int k){
20     pos.loc = n, pos.step = 0;
21     vis[n] = 1;
22     que.push(pos);
23     while(!que.empty()){
24         pos = que.front();
25         que.pop();
26         if(pos.loc == k)
27             return pos.step;
28         int loc_mov = pos.loc - 1;
29         if(loc_mov >= 0 && loc_mov < 200005 && vis[loc_mov] == 0){
30             Push(loc_mov, pos.step);
31         }
32         loc_mov = pos.loc + 1;
33         if(loc_mov >= 0 && loc_mov < 200005 && vis[loc_mov] == 0){
34             Push(loc_mov, pos.step);
35         }
36         loc_mov = 2 * pos.loc;
37         if(loc_mov >= 0 && loc_mov < 200005 && vis[loc_mov] == 0){
38             Push(loc_mov, pos.step);
39         }
40     }
41     return -1;
42 }
43 
44 int main(){
45     int n, k;
46     cin >> n >> k;
47     cout << bfs(n, k) << endl;
48 }

 

Catch That Cow


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