BFS应用：POJ3278 抓牛

题目描述：

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?

输入：Line 1: Two space-separated integers: N and K

输出：Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

输入样例：

5 17

输出样例：

4

解题思路：

        农民有3种移动方式，但如果想向后移动就只有一种移动方式，也就是说明如果农民的位置在牛的位置之前或相同（N>=K）农民就只能采用一步步回退的方式；

        如果牛在农民前面则3种移动方式都能用的上，想求出移动的步数可以BFS的方式逐层扩散，直到遍历到牛的位置，遍历的层数就是最短的路径。

解题步骤：

        1、若N>=K，直接输出N-K；

        2、若N<K，用BFS，定义一个步数数组step，下标为结点的值，内容为该结点到N的步数；

        3、设出发顶点为N，step[N]为0，将其入队后开始遍历；若队头元素为u，三个邻接点为u+1、u-1、2*u，且邻接点的步数比结点的步数+1。

        4、有队头元素 = k时，输出此元素的步数。

代码实现：

#include <iostream>
using namespace std;
#include <queue>
#define maxn 100001

int n, k;
bool vis[maxn];
int step[maxn];

void BFS()
{
    int u = 0, v = 0;
    step[n] = 0;
    queue<int> q;
    q.push(n);
    vis[n] = true;
    while (!q.empty())
    {
        u = q.front();
        q.pop();
        if (u == k)
        {
            cout << step[u] << endl;
            return;
        }
        v = u - 1;
        if (v >= 0 && v <= 100000 && !vis[v])
        {
            step[v] = step[u] + 1;
            vis[v] = true;
            q.push(v);
        }
        v = u + 1;
        if (v >= 0 && v <= 100000 && !vis[v])
        {
            step[v] = step[u] + 1;
            vis[v] = true;
            q.push(v);
        }
        v = 2 * u;
        if(v >= 0 && v <= 100000 && !vis[v])
        {
            step[v] = step[u] + 1;
            vis[v] = true;
            q.push(v);
        }
    }
}

int main()
{
    cin >> n >> k;
    if (n >= k)
    {
        cout << n - k << endl;
    }
    else
    {
        BFS();
    }

    return 0;
}

标签：抓牛,cow,point,POJ3278,move,BFS,Farmer,John,步数
来源： https://blog.csdn.net/qq_48988844/article/details/120350458