Catch That Cow POJ - 3278（一维bfs）

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
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.

#include 
#include 
#include 
using namespace std;

const int maxn = 100000 + 5;
int dis[maxn];
int vis[maxn];
queue<int>Q;

void bfs()
{
	while (!Q.empty())
	{
		int now = Q.front();
		Q.pop();
		for (int i = 0; i < 3; i++)
		{
			int nex;
			if (i == 0)
				nex = now - 1;
			else if (i == 1)
				nex = now + 1;
			else
				nex = now * 2;
			if (nex >= 0 && nex <= 100000 && vis[nex] == 0)
			{
				vis[nex] = 1;
				dis[nex] = 1 + dis[now];
				Q.push(nex);
			}
		}
	}
}

int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		while (!Q.empty())
		{
			Q.pop();
		}
		if (n >= m)
		{
			printf("%d\n", n - m);
			continue;
		}
		memset(vis, 0, sizeof(vis));
		memset(dis, 0, sizeof(dis));
		vis[n] = 1;
		Q.push(n);
		bfs();
		printf("%d\n", dis[m]);
	}
	return 0;
}