POJ 3278 Catch That Cow （有思路有细节的bfs）

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

别看题简单，满满的坑啊！！你初始在n，先让你移动到k，有三种移动，—1，+1，*2。问你几步能到k。
最开始无脑搜，直接T了。后来发现bfs虽然是有点暴力的感觉但是还是要有思维在里面的！如果n>k，n只能一点点减到k。还有数组开大点，以防*2后越界。
代码如下：

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <queue>
 using namespace std;
 int n,k,ans;
 bool vis[];
 struct node
 {
     int pos,step;
 };
 bool check (int x)
 {
     if (vis[x])
     return ;
     if (x<||x>)
     return ;
     else
     return ;
 }
 void bfs (node now)
 {

     queue<node>q;
     q.push(now);
     while (!q.empty())
     {
         node temp=q.front();
         q.pop();
         if (check(temp.pos-))
         {
             node x;
             x.pos=temp.pos-;
             x.step=temp.step+;
             vis[x.pos]=;
             if (x.pos==k)
             {
                 ans=x.step;
                 return ;
             }
             q.push(x);
         }
         if (check(temp.pos+)&&temp.pos<k)
         {
             node x;
             x.pos=temp.pos+;
             x.step=temp.step+;
             vis[x.pos]=;
             if (x.pos==k)
             {
                 ans=x.step;
                 return ;
             }
             q.push(x);
         }
         if (check(temp.pos*)&&temp.pos<k)
         {
             node x;
             x.pos=temp.pos*;
             x.step=temp.step+;
             vis[x.pos]=;
             if (x.pos==k)
             {
                 ans=x.step;
                 return ;
             }
             q.push(x);
         }
     }
 }
 int main()
 {
     //freopen("de.txt","r",stdin);
     while (~scanf("%d%d",&n,&k))
     {
         memset(vis,false,sizeof vis);
         node now;
         now.pos=n;
         now.step=;
         ans=;
         if (n>=k)
         printf("%d\n",n-k);
         else
         {
             bfs(now);
             printf("%d\n",ans);
         }
     }
     return ;
 }