POJ 3278 Catch That Cow【广度优先搜索】

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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本文介绍了一个有趣的算法问题——如何利用最少的时间在数轴上捕捉一只静止不动的牛。通过广度优先搜索策略，探讨了两种移动方式：步行与瞬移，并提供了完整的C++实现代码。

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Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 111610		Accepted: 34874

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

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

题意：在n位置的农夫想抓住在k位置的牛，他们都在一条直线上，假设牛的位置一直保持不变，农夫有两种方法走到牛所在的位置。一是：步行，每分钟只走一步，即设农夫位置为x，向牛所在的方向走，则 x+1，向其反方向走 x-1。二是：传送，在一分钟内，每传送一次都是传送前位移的两倍，即2*x。其抓到牛所花时间最短为多少。

解题思路：该题要用到广度优先搜索。本题其位移变换有三种，即x+1,x-1,x*2。我们让每一次都执行 x=x-1,x=x+1,x=2*x。直到到达牛的所在位置的点为止。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=100005;
int vis[MAXN];
queue<int>q;
int X[3]={0,-1,1};
int bfs(int n,int k)
{
    vis[n]=1;
    q.push(n);
    while(!q.empty())
    {
        int head=q.front();
        q.pop();
        if(head==k)///到达牛所在的位置，就结束
        {
            return vis[head];
        }
        else       
        {
            for(int i=1;i<3;i++)
            {
                int tt=head+X[i];
                if(tt>=0&&tt<=1e5&&vis[tt]==0)///如果当前位置没被访问过就入队
                {
                    vis[tt]=vis[head]+1;
                    q.push(tt);
                }
                int mm=2*head;
                if(mm>=0&&mm<=1e5&&vis[mm]==0)
                {
                    vis[mm]=vis[head]+1;
                    q.push(mm);
                }
            }
        }
    }

}
int main()
{
    int n,k;
    while(~scanf("%d %d",&n,&k))
    {
        memset(vis,0,sizeof(vis));
       while(!q.empty()) q.pop();///注意每一次都要判断队列是否为空
       int  ans= bfs(n,k);
       printf("%d\n",ans-1);///上面模块中起点也计算在其中了，所以这里要减去1.
        }
        return 0;
}