hdu 2717 Catch That Cow(广搜bfs)

题目链接：http://i.cnblogs.com/EditPosts.aspx?opt=1

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7909    Accepted Submission(s):
2498

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.

 

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题目大意：在一个数轴上有n和k，农夫在n这个位置，奶牛在k那个位置，农夫要抓住奶牛，有两种方法：1、walking即农夫可以走x+1的位置和x-1的位置。

2、teleporting即每分钟可以走到2*x的位置。利用这两种方法，找到最快几步可以到达！

 

解题思路：其实就这三种情况，本来没打算用搜索的，不过发现好多东西都可以灵活运用！我们吧第一种方法看成两个方向，用dir[2]={1,-1,}表示，然后进行广搜！第二种方法自然就要特殊判断一下了！有一个地方要特殊判断一下，就是因为数据比较大，vis[ss.x]可能会超范围导致RE，所以都加一行判断使其保证在题目范围中。

 

详见代码。

 

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 #include <queue>

 using namespace std;

 int dir[]= {,-};

 struct node
 {
     int x,step;
 } s,ss;

 int bfs(int n,int k)
 {
     queue<node>q,qq;
     s.x=n;
     s.step=;
     int vis[]= {};
     q.push(s);
     while (!q.empty())
     {
         s=q.front();
         q.pop();
         if (s.x==k)
             return s.step;
         for (int i=; i<; i++)
         {
             ss.x=s.x+dir[i];
             ss.step=s.step+;
             if (ss.x>=&&ss.x<=)
                 if (!vis[ss.x])
                 {
                     vis[ss.x]=;
                     q.push(ss);
                 }
         }
         ss.x=s.x*;
         ss.step=s.step+;
         if (ss.x>=&&ss.x<=)
         {
             if (!vis[ss.x])
             {
                 vis[ss.x]=;
                 q.push(ss);
             }
         }
     }
     return ;
 }

 int main ()
 {
     int n,k;
     while (~scanf("%d%d",&n,&k))
     {
         printf ("%d\n",bfs(n,k));
     }
     return ;
 }