POJ-3278 Catch That Cow （C++）

POJ-3278抓牛算法解析

原创已于 2022-11-18 09:56:18 修改 · 271 阅读

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#算法 #c++ #数据结构 #bfs #广度搜索

于 2021-03-18 02:14:02 首次发布

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本文解析了POJ-3278抓牛问题的算法实现，通过使用C++语言进行广度优先搜索（BFS），有效地解决了农民追赶逃逸奶牛的问题。文中详细介绍了算法步骤，并提供了完整的代码示例。

POJ-3278 Catch That Cow （C++）

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.

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.

my analysis

This problem is essentially given n, through three kinds of specified operation to get k, and to ensure the minimum number of operations, direct BFS breadth first search can be, similar to Luo Gu’s “strange elevator” problem.

solution（C++）

#include<iostream>
#include<queue>
using namespace std;
int N, K;
bool inq[200000] = { false };
int step[200000] = { 0 };
int BFS()
{
	queue<int> q;
	q.push(N);
	while (!q.empty())
	{
		int top = q.front();
		q.pop();
		if (top == K)
			return step[top];
		if (top - 1 >= 0 && inq[top - 1] == false)
		{
			q.push(top - 1);
			step[top - 1] = step[top] + 1;
			inq[top - 1] = true;
		}
		if (top + 1 <= 100000 && inq[top + 1] == false)
		{
			q.push(top + 1);
			step[top + 1] = step[top] + 1;
			inq[top + 1] = true;
		}
		if (top * 2 <= 100000 && inq[top * 2] == false)
		{
			q.push(top * 2);
			step[top * 2] = step[top] + 1;
			inq[top * 2] = true;
		}
	}
	return -1;
}
int main()
{
	cin >> N >> K;
	cout << BFS() << endl;
}