Catch That Cow

最新推荐文章于 2021-02-24 12:03:50 发布

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本文介绍了一个有趣的算法问题：农民约翰如何通过步行与瞬间传送的方式，在最短时间内捕捉到静止不动的逃逸奶牛。该问题涉及到图遍历算法中的广度优先搜索（BFS），并提供了一个具体的实现示例。

M - Catch That Cow

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.

#include<iostream>
#include<cstring>
#include<algorithm>
#include<stdio.h>
#include<queue>
typedef long long ll;
#define maxn 100010
using namespace std;
#define cle(n)    memset(n,0,sizeof(n))
int n,k;
int vis[2*maxn];
struct Node
{
    int x,step;
    Node(){}
    Node(int x,int step):x(x),step(step){}
};
int bfs()
{
    queue<Node>que;
    que.push(Node(n,0));

    vis[n]=1;
    while(que.size())
    {
        Node node=que.front();
         que.pop();
        for(int i=0;i<3;i++)
        {
            int nx=0;
            if(i==0)
                 nx=node.x-1;
            if(i==1)
                nx=node.x+1;
            if(i==2)
                nx=node.x*2;

            if(nx==k)
            {
               cout<<node.step+1<<endl; return 0;
            }
            if(nx<0||nx>100000)
                continue;
            if(!vis[nx])
            {
                vis[nx]=1;
                que.push(Node(nx,node.step+1));
            }

        }

    }

}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
     cle(vis);
   if(n<k)
      bfs();
      if(n==k)
        cout<<0<<endl;
      if(n>k)
        cout<<n-k<<endl;

    }


    return 0;
}