poj3278

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最新推荐文章于 2025-03-30 16:45:00 发布

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本文探讨了农民利用行走和瞬移两种方式追捕逃逸奶牛的最短时间问题，通过路径优化算法求解最优策略，实现快速定位目标。

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题目名称：Catch That Cow

题目链接：http://poj.org/problem?id=3278

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.

题意:一个农民追牛，每次可以走-1或1步和走2*x步，问追上的最短时间

思路：

1. X往左走当X小于0的时候，2*X不可能大于0了，所以X－1的极限是X-1 >= 0。
2. X往右走当X大于K的时候，不管是2*X还是X－1只会增加到达K的步数，所以X＋1的极限是X+1 <= k。
3. 当X>K的时候，2*X也只会增加X到达点K的步数，所以2*X的极限是X<K。

代码如下：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct Node
{
    int x,sum;
};
Node tmp,now;
int n,k;
bool vis[1000005];
queue<Node> q;
void bfs()
{
    while(!q.empty())
        q.pop();
    q.push(tmp);
    while(!q.empty())
    {
        Node now=q.front();
        q.pop();
        if(now.x==k)
        {
            printf("%d\n",now.sum);
            return;
        }
        if(now.x-1>=0&&!vis[now.x-1])
        {
            vis[now.x-1]=true;
            tmp.x=now.x-1;
            tmp.sum=now.sum+1;
            q.push(tmp);
        }
        if(now.x+1<=k&&!vis[now.x+1])
        {
            vis[now.x+1]=true;
            tmp.x=now.x+1;
            tmp.sum=now.sum+1;
            q.push(tmp);
        }
        if(now.x<=k&&now.x*2<=100000&&!vis[now.x*2])
        {
            vis[now.x*2]=true;
            tmp.x=now.x*2;
            tmp.sum=now.sum+1;
            q.push(tmp);
        }
    }

}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(vis,false,sizeof(vis));
        tmp.x=n;
        tmp.sum=0;
        bfs();
    }
    return 0;
}