POJ - 3278 Catch That Cow

最新推荐文章于 2025-03-30 16:45:00 发布

帆帆帆帆帆丶

最新推荐文章于 2025-03-30 16:45:00 发布

阅读量176

点赞数

分类专栏：搜索

本文链接：https://blog.youkuaiyun.com/qq_36344542/article/details/79488536

版权

搜索专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种使用广度优先搜索(BFS)算法来解决寻找逃逸奶牛最短路径的问题。Farmer John从点N出发，在同一数轴上找到位于点K的奶牛。通过行走或传送的方式，算法详细展示了如何计算达到目标位置所需的最少时间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

据说有剪枝后续更新

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
bool vis[100001];
int step[100001];
queue<int>q;
int bfs(int n,int k)
{
    int head,next;
    q.push(n);//n入队
    step[n]=0;
    vis[n]=true;//标记已访问
    while(!q.empty())
    {
        head=q.front();
        q.pop();//取队首并将其弹出
        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<0||next>=100001)
                continue;//判断出界
            if(!vis[next])//如果没有被访问
            {
                q.push(next);//入队
                step[next]=step[head]+1;//步数加1
                vis[next]=true;
            }
            if(next==k)
                return step[next];//结束
        }
    }
}
int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        memset(step,0,sizeof(step));
        memset(vis,false,sizeof(vis));
        while(!q.empty())
            q.pop();
        if(n>=k)
            printf("%d\n",n-k);
        else
        {
            int ans;
            ans=bfs(n,k);
            printf("%d\n",ans);
        }
    }
    return 0;
}