hdoj 2717 Catch That Cow

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.

讲解：一农夫追牛，牛很笨，他就在原地等着农夫来抓他，并且这是一条直线，农夫很容易就能找到它的，然而农夫却只有三种选择，退一步，走一步，或者走到当前位置的2倍；于是乎我们可以用搜索来解决；

代码如下：

 #include<algorithm>
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 using namespace std;
 int n,m;
 int map[];
 struct T
 {
     int x,step;
 };
 int dfs(T now)
 {queue< T >q;
     T end;
     q.push(now);
     while(!q.empty())
     {
         end=q.front();
           q.pop();
         map[end.x]=;
     if(end.x==m)
      {return end.step;}
      //如果不是走一步一判断很容易超出内存的
      now.x=end.x+;//前进一步，存入队列；
          if(end.x>= && end.x<= && map[now.x]==)
          {
              now.step=end.step+;
              map[now.x]==;
              q.push(now);
          }
      now.x=end.x-;//后退一步
      if(end.x>= && end.x<= && map[now.x]==)
      {
          now.step=end.step+;
          map[now.x]==;
          q.push(now);
      }
      now.x=end.x*;//前进2倍的位置
      if(end.x>= && end.x<= && map[now.x]==)
      {
          now.step=end.step+;
          map[now.x]==;
          q.push(now);
      }
     }
     return ;
 }
 int main()
 {
     T now;
     while(cin>>n>>m)
     {
         if(n>=m)//如果n大于m则只能后退了；
        {
            cout<<n-m<<endl;continue;
        }
        else
        {
         memset(map,,sizeof(map));
         now.x=n;now.step=;
         int mm=dfs(now);
         cout<<mm<<endl;
        }
     }
     return ;
 }