[ACM] hdu 2717 Catch That Cow (BFS）

最新推荐文章于 2023-03-24 16:33:19 发布

原创最新推荐文章于 2023-03-24 16:33:19 发布 · 2.9k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#BFS #ACM

ACM题目同时被 3 个专栏收录

340 篇文章

订阅专栏

ACM之路

286 篇文章

订阅专栏

[ACM]_搜索

23 篇文章

订阅专栏

本文介绍了一个有趣的算法问题——如何让农夫约翰以最短时间抓到逃跑的奶牛。通过广度优先搜索（BFS）算法，我们解决了这个问题，并详细解释了实现过程及注意事项。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Source

USACO 2007 Open Silver

解题思路：

从位置n到位置k根据规则最少走几步可以到达，规则是从n可以走到n-1,可以走到n+1,也可以走到n*2.用bfs广搜来做，vis[]用来记录是否访问，hash[]用来记录走的步数。

遇到的问题有数组越界，一定要先判断n*2,n-1,n+1是否越界，还有问题就是步数的保存，因为很多节点都是同一个步数下的状态，因为每一步根据规则可以衍生出三种状态。所以用hash[ q.front() 下一个状态]=hash[q.front()]+1.用这种方法来记录步数，最后直接输出hash[k]就可以了。

代码：

#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
const int len=100000;
int hash[len+10];//hash[i]用来保存走到位置i时是第几步
bool vis[len+10];
int n,k;

void bfs()
{
    memset(vis,0,sizeof(vis));
    queue<int>q;
    q.push(n);
    vis[n]=1;
    hash[n]=0;
    while(!q.empty())
    {
        int temp=q.front();
        q.pop();
        if(temp+1==k||temp-1==k||temp*2==k)//找到
        {
            hash[k]=hash[temp]+1;
            break;
        }
        if(temp-1>=0&&!vis[temp-1])
        {
            q.push(temp-1);
            vis[temp-1]=1;
            hash[temp-1]=hash[temp]+1;
        }
        if(temp+1<=len&&!vis[temp+1])
        {
            q.push(temp+1);
            vis[temp+1]=1;
            hash[temp+1]=hash[temp]+1;
        }
        if(temp*2<=len&&!vis[temp*2])
        {
            q.push(temp*2);
            vis[temp*2]=1;
            hash[temp*2]=hash[temp]+1;
        }
    }
}

int main()
{
    while(cin>>n>>k)
    {
        if(n==k)
        {
            cout<<0<<endl;
            continue;
        }
        else
            bfs();
        cout<<hash[k]<<endl;
    }
    return 0;
}