Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 33021 Accepted Submission(s): 8956
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?
* 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 #define scanf scanf_s 7 const int N = 1000000; 8 9 struct node { 10 int x, step; 11 }; 12 int n, k; 13 bool vis[N+10]; 14 15 int check(int x) { 16 if (x < 0 || x >= N || vis[x]) return 0; 17 else return 1; 18 } 19 20 21 int bfs(int x) { 22 memset(vis, 0, sizeof(vis)); 23 queue<node> q; 24 node a; 25 a.x = x; a.step = 0; vis[a.x] = 1; 26 q.push(a); 27 while (!q.empty()) { 28 node a = q.front(); 29 q.pop(); 30 if (a.x == k) return a.step; 31 node nxt = a; 32 nxt.x = a.x + 1; 33 if (check(nxt.x)) { 34 nxt.step = a.step + 1; 35 vis[nxt.x] = 1; 36 q.push(nxt); 37 } 38 nxt.x = a.x - 1; 39 if (check(nxt.x)) { 40 nxt.step = a.step + 1; 41 vis[nxt.x] = 1; 42 q.push(nxt); 43 } 44 nxt.x = a.x * 2; 45 if (check(nxt.x)) { 46 nxt.step = a.step + 1; 47 vis[nxt.x] = 1; 48 q.push(nxt); 49 } 50 } 51 return -1; 52 53 } 54 55 int main() { 56 int ans; 57 while (~scanf("%d%d", &n, &k)) 58 { 59 ans = bfs(n); 60 printf("%d ", ans); 61 } 62 63 64 return 0; 65 }