POJ 3278 Catch That Cow

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 30437		Accepted: 9390

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

4

这个看完了我顿时沸腾了起来，写的太好了，bfs与队列的完美结合，谢谢作者。

#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;
}