HDU 2717 Catch That Cow

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11487    Accepted Submission(s): 3560

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. 
    
    
    
    
     
   
   
   
   

 

题意：给一个农夫的坐标和奶牛的坐标   农夫可以走的路径有三种  x-1,x+1,2*x;（不过这题没说农夫的坐标一定小于奶牛的，试了一下考虑不考虑都能A）

             是一道简单的搜索bfs，才做没往搜索想，一直想着暴力  还是做的题少学得算法少 不管做什么题开始总想暴力 根本想不到用什么算法  只有仔细思考才能看出来

代码：

         

#include<stdio.h>
#include<string.h>
#include<queue>
#define INF 100005
using namespace std;
int n,m;
int d[200005];
int bfs(int n,int m)
{
    queue<int>q;
    q.push(n);
    d[n]=0;
    while(!q.empty())
    {
        int p=q.front();
        q.pop();
        if(p==m)
        {
            break;
        }
        for(int i=0; i<3; i++)
        {
            int nx;
            if(i==0)
            {
                nx=p+1;
            }
            if(i==1)
            {
               nx=p-1;
            }
            if(i==2)
            {
                 nx=p*2;
            }
            if(nx>=0&&nx<=INF&&d[nx]==INF)
            {
                q.push(nx);
                d[nx]=d[p]+1;
            }
        }
    }
    return d[m];
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
          for(int i=0;i<=INF;i++)
          {
              d[i]=INF;
          }
           int ans= bfs(n,m);
            printf("%d\n",ans);
    }
}