Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 31637 | Accepted: 9740 |
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 //注意下标不可变为-1，两个数组均要开到20w，为此吃了好几个re 2 //大致题意：n可以执行加一减一乘二，求至少几步到k 3 #include <iostream> 4 #include <queue> 5 #include <cstring> 6 #include <cstdlib> 7 using namespace std; 8 const int N = 100000; 9 int a[200005]; 10 queue <int > q; 11 int vis[N*2+5]; 12 void bfs(int n,int k) 13 { 14 vis[n] = true; 15 q.push(n); 16 while(!q.empty()) 17 { 18 int head = q.front(); 19 q.pop(); 20 if(head==k) 21 { 22 return ; 23 } 24 else 25 {//原来没加大于0的条件，结果re 26 if(!vis[head-1]&&head<=N&&head>=1)//禁止下标变为-1 27 { 28 vis[head-1] = true; 29 a[head-1]+=a[head]+1; 30 q.push(head-1); 31 } 32 if(!vis[head+1]&&head<N&&head>=0) 33 { 34 vis[head+1] = true; 35 a[head+1]+=a[head]+1; 36 q.push(head+1); 37 } 38 if(!vis[head*2]&&head<=50000&&head>=0) 39 { 40 vis[head*2] = true; 41 a[head*2]+=a[head]+1; 42 q.push(head*2); 43 } 44 } 45 } 46 } 47 int main() 48 { 49 int n,k; 50 cin>>n>>k; 51 memset(vis,false,sizeof(vis)); 52 memset(a,0,sizeof(a)); 53 while(!q.empty()) 54 q.pop(); 55 bfs(n,k); 56 if(n>=k) 57 cout<<n-k<<endl; 58 else 59 cout<<a[k]<<endl; 60 system("pause"); 61 return 0; 62 }
