HDU 2717 Catch That Cow （bfs）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2717

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

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.

 

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：

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <stack>
 #include <queue>
 using namespace std;
 int f[];  //记录步数
 void bfs(int n,int k)
 {
     int a[];  //位置数组
     memset(f,,sizeof(f));
     queue <int > q;
     q.push(n);
     f[q.front()] = ;
     if (n == k)
         return ;
     while (!q.empty())
     {
         int t = q.front();
         int x = f[t];
         a[] = t-;  //三个位置
         a[] = t+;
         a[] = t*;
         for (int i = ; i < ; i ++)
         {
             if (a[i] >=  && a[i] <  && f[a[i]]==)  //注意这里的边界值
             {
                 q.push(a[i]);
                 f[a[i]] = x+;  //在上一步的基础上加1
             }
             if (a[i] == k)
                 return ;
         }
         q.pop();
     }
 }
 int main ()
 {
     int i;
     int n,k;
     while (~scanf("%d%d",&n,&k))
     {
         dfs(n,k);
         printf("%d\n",f[k]-); //减去农夫本来在的位置那一步
     }
     return ;
 }