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最新推荐文章于 2025-03-30 16:45:00 发布

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最新推荐文章于 2025-03-30 16:45:00 发布

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本文介绍了一个经典的BFS（广度优先搜索）算法问题——CatchThatCow。问题描述了农夫约翰如何利用步行和瞬移两种方式在最短时间内追上静止不动的逃逸奶牛。通过分析并给出C++实现代码，展示了如何高效解决此类搜索问题。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37800		Accepted: 11724

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

一道bfs的大水题，下面是代码l:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>

using namespace std;

const int maxn=100005;

int step[maxn];
bool vis[maxn];
queue<int> q;
int n,k;
int next,head;

int bfs(){
    q.push(n);
    step[n]=0;
    vis[n]=1;
    while(!q.empty()){
         head=q.front();
         q.pop();
         for(int i=0;i<3;i++){
             if(i==0) next=head-1;
             else if(i==1) next=head+1;
             else next=head*2;
             if(next>maxn||next<0) continue;
             if(!vis[next]){
                  q.push(next);
                  vis[next]=1;
                  step[next]=step[head]+1;
             }
             if(next==k)
                 return step[next];
         }
    }
    return -1;
}

int main(){
   cin>>n>>k;
   memset(vis,0,sizeof(vis));
    if(n>=k)//开始写的时候忘记写了
       cout<<n-k<<endl;
    else
       printf("%d\n",bfs());
       return 0;
}