POJ3278,Catch That Cow,BFS...

最新推荐文章于 2024-07-24 11:36:30 发布

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最新推荐文章于 2024-07-24 11:36:30 发布

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农夫John在数轴上追赶逃逸的牛，通过行走或瞬移方式移动，目标是找到最快到达牛位置的时间。此题通过广度优先搜索（BFS）解决，强调了算法在解决问题中的应用。

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Catch That Cow

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.

分析：

农夫追赶牛，只有两种走的方式，其实是三种...往左右走，或者下标变成两倍。入门级的bfs，关键是好久都是用n个0结尾的题，忘了判断EOF，

贡献了好几个OLE

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#define MAX 200030
using namespace std;
typedef struct node
{
    int x,step;
}Node;
int n,k;
int vis[MAX+5];
Node que[MAX+5];
void bfs()
{
    int front=0,rear=0;
    que[rear].x=n;
    que[rear++].step=0;
    vis[n]=1;
    while(front<rear)
    {
        Node tmp=que[front++];
        if(tmp.x==k)
        {
            printf("%d\n",tmp.step);
            return;
        }
        if(tmp.x-1>=0&&!vis[tmp.x-1])
        {
            vis[tmp.x-1]=1;
            que[rear].x=tmp.x-1;
            que[rear++].step=tmp.step+1;
        }
        if(tmp.x<=k&&!vis[tmp.x+1])
        {
            vis[tmp.x+1]=1;
            que[rear].x=tmp.x+1;
            que[rear++].step=tmp.step+1;
        }
        if(tmp.x<=k&&!vis[2*tmp.x])
        {
            vis[2*tmp.x]=1;
            que[rear].x=2*tmp.x;
            que[rear++].step=tmp.step+1;
        }
    }
}
int main()
{
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        bfs();
    }
    return 0;
}