POJ-3278 Catch that cow

最新推荐文章于 2024-07-24 11:36:30 发布

luer9

最新推荐文章于 2024-07-24 11:36:30 发布

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分类专栏： BFS

本文链接：https://blog.youkuaiyun.com/qq_40046426/article/details/79166203

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本文介绍了一种算法，用于解决农夫约翰如何最短时间捕捉到静止不动的逃逸奶牛的问题。通过广度优先搜索（BFS），算法详细展示了如何从起始位置出发，利用步行和传送两种方式，计算到达目标位置所需的最少时间。

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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.

代码如下：

#include <iostream>
#include <stack>
#include <math.h>
#include <string>
#include <queue>
#include <cstdio>
using namespace std;
queue<int>map;
int vis[100005];//记录位置
int ans[100005];//记录时间
int n,k;//n是起点  k是终点
int bfs(int pos,int end)
{
    map.push(pos);//把农夫的位置放进队列中
    ans[pos]=0;//记录开始时间点为0
    while(!map.empty())
    {
        int t=map.front();//当队列里还有数的时候  取t为队列首的数  然后把它拿出来
        map.pop();
        for(int i=0;i<3;i++)//循环遍历三种情况
        {
            int x;//x是下一步可以到达的值
            if(i==0)
                x=t-1;
            else if(i==1)
                x=t+1;
            else if(i==2)
                x=2*t;
            if(x>2*end||x<0||x>100001)//判断是否越界
                continue;
            if(!vis[x])//如果x那一格没有被访问过
            {
                vis[x]=1;//将vis那一格标记为1表示已经走过一次
                map.push(x);//把x放进队列
                ans[x]=ans[t]+1;//到x的时间等于到t的时间加上1
            }
            if(x==end)
                return ans[k];
        }
    }
    //return pos;
}
int main()
{
    scanf("%d %d",&n,&k);
    memset(vis,0,sizeof(vis));
    memset(ans, 0, sizeof(ans));
    while(!map.empty())
        map.pop();
    if(n>=k)
        printf("%d\n",n-k);//如果农夫在牛前面  只要每次往后走一步 走n-k次就行了
    else
        printf("%d\n",bfs(n,k));//否则就进入BFS广搜
    return 0;
}