Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12615 Accepted Submission(s): 3902

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?

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

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.

题目大意：在一条笔直的道路上，有一个农夫在A位置，一头牛在B位置，农夫一次可以搜索三个位置（例如农夫在x位置，他可以搜索x-1、x+1、x*2）问农夫至少需要几步能够找到牛。

解题思路：用广搜，定义一个位置数组，每次搜索三个位置即可。

AC代码：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <stack>
 6 #include <queue>
 7 using namespace std;
 8 int f[200020];  //记录步数 
 9 void bfs(int n,int k)
10 {
11     int a[3];  //位置数组 
12     memset(f,0,sizeof(f));
13     queue <int > q;
14     q.push(n);
15     f[q.front()] = 1;
16     if (n == k)
17         return ;
18     while (!q.empty())
19     {
20         int t = q.front();
21         int x = f[t];
22         a[0] = t-1;  //三个位置 
23         a[1] = t+1;
24         a[2] = t*2;
25         for (int i = 0; i < 3; i ++)
26         {
27             if (a[i] >= 0 && a[i] < 200001 && f[a[i]]==0)  //注意这里的边界值 
28             {
29                 q.push(a[i]);
30                 f[a[i]] = x+1;  //在上一步的基础上加1 
31             }
32             if (a[i] == k)
33                 return ;
34         }
35         q.pop();
36     }
37 }
38 int main ()
39 {
40     int i;
41     int n,k;
42     while (~scanf("%d%d",&n,&k))
43     {
44         dfs(n,k);
45         printf("%d
",f[k]-1); //减去农夫本来在的位置那一步 
46     }
47     return 0;
48 }