HDU 2717： Catch that cow （BSF）

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#HDU 2717 Catch that #BSF

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本文介绍了一道经典的算法题目，农夫通过步行和瞬移的方式追赶静止不动的牛，旨在寻找最短路径。使用广度优先搜索(BFS)算法解决此问题，并提供了完整的C++实现代码。

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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

农夫需要找到他的牛，农夫现在所在位置是N，牛在K点，假设牛是不动的，农夫有三种走法：

向前走一步（N+1）
向后走一步（N-1）
跳到当前位置两倍处（2*N）

问，最少走多少次才能找到他的牛。

解题思路：把所有的情况都列举一遍，最先找到的，就是用次最少的。

具体代码：

#include<cstdio>
#include<cstring>
#include<queue>
const int M=300010;
using namespace std;
struct pon{
    int x,ans;
};
int vis[M],n,k;
int bfs()
{
    pon p,pp;
    memset(vis,0,sizeof(vis));
    queue<pon> q;
    p.ans=0;p.x=n;
    q.push(p);
    vis[p.x]=1;
    while(!q.empty())
    {
        p=q.front();
        q.pop();
        if(p.x==k)
        {
            return p.ans;
        }
        if(p.x+1<M&&!vis[p.x+1])
        {
            pp.x=p.x+1;
            pp.ans=p.ans+1;
            q.push(pp);
            vis[p.x]=1;
        }
        if(p.x-1>=0&&p.x-1<M&&!vis[p.x-1])
        {
            pp.x=p.x-1;pp.ans=p.ans+1;
            q.push(pp);
            vis[p.x]=1;
        }
        if(p.x*2<M&&!vis[p.x*2])
        {
            pp.x=p.x*2;pp.ans=p.ans+1;
            q.push(pp);
            vis[p.x]=1;
        }
    }
}
int main()
{
    int ans;
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        ans=bfs();
        printf("%d\n",ans);


    }
    return 0;
}