POJ 3278 Catch That Cow

Posted 2020-08-21 IKnowYou

POJ: https://i.cnblogs.com/EditPosts.aspx?opt=1

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 80291		Accepted: 25297

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.

 1 #include <iostream>
 2 #include <queue>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 using namespace std;
 7 
 8 const int N = 100009;
 9 
10 int main() {
11     int visited[N];
12     int n, k;
13     while (cin>>n>>k) {
14         queue<int>q;
15         memset(visited, 0, sizeof(visited));
16         q.push(n);
17         while (!q.empty()) {
18             int tmp = q.front();
19             q.pop();
20             if (tmp == k)
21                 break;
22             if (tmp-1 >=0 && visited[tmp - 1] == 0) {
23                 visited[tmp - 1] = visited[tmp] + 1;
24                 q.push(tmp - 1);
25             }
26             if (tmp+1 <= k&&visited[tmp + 1] == 0) {
27                 visited[tmp + 1] = visited[tmp] + 1;
28                 q.push(tmp + 1);
29             }
30             if (tmp * 2 < N&&visited[2 * tmp] == 0) {
31                 visited[2 * tmp] = visited[tmp] + 1;
32                 q.push(2 * tmp);
33             }
34         }
35         cout << visited[k] << endl;
36     }
37     return 0;
38 
39 }