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POJ 3278 Catch That Cow C++版

POJ题解 Slyar 4529浏览 3评论

文章作者:姜南(Slyar) 文章来源:Slyar Home (www.slyar.com) 转载请注明,谢谢合作。

最基础的广搜题,拿<queue>玩一下…

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

Output

For each k given in the input, print one line containing ak to the output.

Sample Input

5 17

Sample Output

4

Slyar:USACO的题,给出2个数n和k,问从n经过+1或者-1或者*2能到达k的最小步数。分3个方向BFS,找到后树的深度就是最小步数了。注意n可以比k大,这时只有-1一种办法可以从n到达k,直接减就行了,还有要注意边界的判断…

#include 
#include 
#define SIZE 100001

using namespace std;

queue x;
bool visited[SIZE];
int step[SIZE];

int bfs(int n, int k)
{
	int head, next;
	//起始节点入队
	x.push(n);
	//标记n已访问 
	visited[n] = true;
	//起始步数为0 
	step[n] = 0;
	//队列非空时 
	while (!x.empty())
	{
		//取出队头 
		head = x.front();
		//弹出队头 
		x.pop();
		//3个方向搜索 
		for (int i = 0; i < 3; i++)
		{
			if (i == 0) next = head - 1;
			else if (i == 1) next = head + 1;
			else next = head * 2;
			//越界就不考虑了 
			if (next > SIZE || next < 0) continue;
			//判重 
			if (!visited[next])
			{
				//节点入队 
				x.push(next);
				//步数+1 
				step[next] = step[head] + 1;
				//标记节点已访问 
				visited[next] = true;
			}
			//找到退出 
			if (next == k) return step[next];
		}
	}
}

int main()
{
	int n, k;
	cin >> n >> k;
	if (n >= k)
	{
		cout << n - k << endl;
	}
	else
	{
		cout << bfs(n, k) << endl;
	}
	return 0;
}

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网友最新评论 (3)

  1. 哈哈哈,从你这学到好多东西啊,非常感谢
    蜗牛10年前 (2011-04-10)回复
  2. 好久没看C++了,当初只学了一点皮毛。
    网络人生11年前 (2010-04-18)回复