POJ 3278 Catch That Cow —————— BFS

浅浅的花香味﹌ 2022-05-13 02:24 198阅读 0赞

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Catch That Cow

<table> 
 <tbody> 
  <tr> 
   <td><b>Time Limit:</b> 2000MS</td> 
   <td></td> 
   <td><b>Memory Limit:</b> 65536K</td> 
  </tr> 
  <tr> 
   <td><b>Total Submissions:</b> 120100</td> 
   <td></td> 
   <td><b>Accepted:</b> 37481</td> 
  </tr> 
 </tbody> 
</table>

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

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.

Source

USACO 2007 Open Silver

—-

/*
        2018/9/9
    
        BFS
        x 有三个搜索方向
        x+1;
        x-1;
        x*2;
        优先队列，找最短时间
    */
    
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #include<queue>
    using namespace std;
    const int MAXN=1e6+9;
    int n,k;
    bool vis[MAXN];
    struct node{
        int x,step;
        node(){};
        node(int _x,int _step)
        {
            x=_x; step=_step;
        }
        bool operator < (const node &b)const
        {
            return step==b.step?x<b.x:step>b.step;
        }
    };
    
    int bfs(int x)
    {
        memset(vis,0,sizeof(vis));
        vis[x]=1;
        node e1,e2;
        priority_queue<node> que;
        que.push(node(x,0));
        int ans=0;
        while(que.size())
        {
            e1=que.top();que.pop();
            if(e1.x==k)
            {
                ans=e1.step;
                break;
            }
            if(!vis[e1.x+1])
            {
                vis[e1.x+1]=1;
                que.push(node(e1.x+1,e1.step+1));
            }
            if(e1.x>0 && !vis[e1.x-1])
            {
                vis[e1.x-1]=1;
                que.push(node(e1.x-1,e1.step+1));
            }
            if(e1.x<=200000 && !vis[e1.x*2])
            {
                vis[2*e1.x]=1;
                que.push(node(e1.x*2,e1.step+1));
            }
        }
        return ans;
    }
    
    int main()
    {
        while(~scanf("%d %d",&n,&k))
            printf("%d\n",bfs(n));
        return 0;
    }