POJ3278——Catch That Cow(BFS)

Catch That Cow

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：

　　　　有个农夫在N点，有个牛在K点。农夫有三种移动方式：1）向前一步 2）向后一步 3）从当所在位置X瞬间移动到2*X点

　　　　输出最少进行多少步，农夫能找到牛。(大坑是农夫和牛的活动范围在[0,100000])

解题思路：

　　　　bfs即可。注意农夫的可活动范围

ps:HUD 2717与这个题一样，但是是多组输入输出，请注意！

Code：

 #include<string>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<iostream>
 using namespace std;
 int dis[],vis[],N,K;
 queue<int> q;
 int bfs(int N,int K)
 {
     q.push(N);
     dis[N]=,vis[N]=;
     while (!q.empty())
     {
         int x[],i;
         int front=q.front();
         q.pop();
         x[]=front-,x[]=front+,x[]=front*;
         for (i=; i<=; i++)
         {
             if (x[i]>=&&x[i]<=&&!vis[x[i]])
             {
                 q.push(x[i]),vis[x[i]]=,dis[x[i]]=dis[front]+;
                 if (x[i]==K) return dis[x[i]];
             }
         }
     }
 }
 int main()
 {
     cin>>N>>K;
     if (N==K) cout<<;
     else
         cout<<bfs(N,K);
     return ;
 }