POJ-3278 Catch That Cow（bfs）

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最新推荐文章于 2025-03-30 16:45:00 发布

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农夫约翰得知一头逃犯奶牛的位置，他从点N开始（0 ≤ N ≤ 100,000）在数轴上追赶奶牛，奶牛位于点K（0 ≤ K ≤ 100,000）。约翰有两种移动方式：步行和瞬移。步行可以让他在一分钟内从点X移动到X-1或X+1，瞬移可以让他在一分钟内从点X移动到2X。题目要求找出约翰追捕奶牛所需的最短时间。通过广度优先搜索（BFS）策略，从三个方向（X-1, X+1, 2*X）同时进行搜索，配合动态规划数组记录已到达的点和步数，最终得出约翰需要4分钟才能抓到奶牛。" 52661827,64035,回调函数与消息机制：原理与应用,"['C++', '函数指针', 'Windows消息机制', '编程设计', '异步编程']

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Problem 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.

思路：

向X-1、X+1和2*X进行bfs，开一个整型的dp数组，既进行了是否已经到达过的标记，又统计了步数。

上AC代码：

#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
int n,k;
int dp[100001];
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(dp,-1,sizeof(dp));
        queue<int> que;
        que.push(n);
        dp[n]=0;
        while(!que.empty())
        {
            int now=que.front();
            if(now==k)
            {
                break;
            }
            que.pop();
            if(now-1>=0&&dp[now-1]==-1)
            {
                dp[now-1]=dp[now]+1;
                que.push(now-1);
            }
            if(now+1<=100000&&dp[now+1]==-1)
            {
                dp[now+1]=dp[now]+1;
                que.push(now+1);
            }
            if(2*now<=100000&&dp[2*now]==-1)
            {
                dp[2*now]=dp[now]+1;
                que.push(2*now);
            }
        }
        printf("%d\n",dp[k]);
    }
    return 0;
}