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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 }