Catch That Cow

最新推荐文章于 2022-01-19 03:53:25 发布

三茶先生

最新推荐文章于 2022-01-19 03:53:25 发布

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本文介绍了一个有趣的算法问题——如何让农夫约翰用最短的时间抓到不动的逃跑的奶牛。通过采用广度优先搜索（BFS）算法，并考虑步行与瞬移两种移动方式，文章详细展示了算法的实现过程。

Catch That Cow
Time Limit : 5000/2000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 4 Accepted Submission(s) : 3
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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.
Source
//三种走法，因为是追赶牛，如果跑牛前面去了，回来只能一步一步走，三种走法分别是向前，退后一步，向前当前步*2，三个。直接bfs就行了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 100000
int vist[N+5];
int k,n;
using namespace std;
struct node
{
    int x,c;
};
int  bfs()
{
    node st,ed;
    queue<node>Q;
    while(!Q.empty())
    {
        Q.pop();
    }
    st.x=n,st.c=0;
    vist[n]=1;
    Q.push(st);
    while(!Q.empty())
    {
        st=Q.front();
        Q.pop();
        for(int i=0; i<3; i++)
        {
            if(i==0)
                ed.x=st.x-1;
            else if(i==1)
                ed.x=st.x+1;
            else ed.x=st.x*2;
            ed.c=st.c+1;
            if(ed.x==k)
                return ed.c;
            if(ed.x>=0&&ed.x<=N&&!vist[ed.x])
            {
                vist[ed.x]=1;
                Q.push(ed);
            }
        }
    }
    return 0;
}
int main()
{
    while(scanf("%d%d",&n,&k)==2)
    {
        if(n>=k)
        {
            printf("%d\n",n-k);
            continue;
        }
        memset(vist,0,sizeof(vist));
        printf("%d\n",bfs());
    }
    return 0;
}