POJ 3278-Catch That Cow（BFS-一维广搜）

阳光穿透心脏的1/2处 2022-07-11 11:26 185阅读 0赞

Catch That Cow

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;2000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;82618</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;25955</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][]

## 题目意思： ##

在一维坐标上，奶牛丢在了K位置，农夫从N位置去找牛。

农夫只能有三种走法：①后退一步；②前进一步；③瞬移到二倍N的位置。

每次使用一种走法花费一分钟，求找到牛要花多久。

## 解题思路： ##

BFS，分别对三种走法广搜，记录走到当前位置花费多长时间，直到N==K。

#include <iostream>
    #include <cstring>
    #include <cstdio>
    #include<queue>
    using namespace std;
    const int maxn=3000000;
    int n,k;
    int ans=0;
    bool vis[maxn];//记录是否访问
    int pace[maxn];//记录走过的所有位置
    int head,pos;//当前位置和要走到的位置
    queue <int> q;
    
    int bfs()
    {
        q.push(n);
        while(!q.empty())
        {
            head=q.front();
            q.pop();
            for(int i=0; i<3; ++i)//有三种走法
            {
                if(i==0) pos=head-1;//后退一步
                else if(i==1) pos=head+1;//前进一步
                else pos=head*2;//二倍步数
                if(pos<0||pos>100000) continue;//超过范围
                if(vis[pos]==false)//未访问过就加入队列
                {
                    q.push(pos);
                    pace[pos]=pace[head]+1;//记录到达当前位置走过多少步
                    //cout<<pos<<endl;
                    vis[pos]=true;
                }
                if(pos==k) return pace[pos];//找到奶牛
            }
        }
    }
    
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        while(cin>>n>>k)//农夫、奶牛位置
        {
            memset(vis,false,sizeof(vis));
            if(n>=k) cout<<n-k<<endl;//如果比目标点远，就只能一步步后退挪回来
            else cout<<bfs()<<endl;
        }
        return 0;
    }
    /*
    5 17
    */

[USACO 2007 Open Silver]: http://poj.org/searchproblem?field=source&key=USACO+2007+Open+Silver