题目地址： https://vjudge.net/problem/POJ-3278
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 129558 | Accepted: 40251 |
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<cstring> 3 #include<stdio.h> 4 #include<queue> 5 6 using namespace std; 7 8 #define MAX 100000 9 int N, K; 10 int vis[MAX + 20]; 11 12 struct node 13 { 14 int x; 15 int step; 16 }s; 17 18 void bfs(int x) 19 { 20 memset(vis, 0, sizeof(vis)); 21 s.x = x; 22 s.step = 0; 23 vis[x] = 1; 24 queue<node> Q; 25 Q.push(s); 26 27 node t; 28 while (!Q.empty()) 29 { 30 t = Q.front(); 31 Q.pop(); 32 33 if (t.x == K) 34 { 35 cout << t.step << endl; 36 return; 37 } 38 39 if (t.x + 1 <= MAX && !vis[t.x + 1]) 40 { 41 s.x = t.x + 1; 42 s.step = t.step + 1; 43 vis[s.x] = 1; 44 Q.push(s); 45 } 46 if (t.x - 1 >= 0 && !vis[t.x - 1]) 47 { 48 s.x = t.x - 1; 49 s.step = t.step + 1; 50 vis[s.x] = 1; 51 Q.push(s); 52 } 53 if (t.x * 2 <= MAX && !vis[t.x * 2]) 54 { 55 s.x = t.x * 2; 56 s.step = t.step + 1; 57 vis[s.x] = 1; 58 Q.push(s); 59 } 60 } 61 return; 62 } 63 64 int main() 65 { 66 while(cin >> N >> K) 67 bfs(N); 68 69 70 return 0; 71 }