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 }