HDU 2717 Catch That Cow (BFS)

Problem 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

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;
#define N 100010
queue q;
int step[N];
int bfs(int n, int k)
{
	while (!q.empty()) q.pop();
	q.push(n); step[n] = 0;
	while (q.front() != k)
	{
		int from = q.front(); q.pop();
		int to = from - 1;
		if (to >= 0 && step[to] == -1)
		{
			q.push(to);
			step[to] = step[from] + 1;
		}
		to = from + 1;
		if (to < N && step[to] == -1)
		{
			q.push(to);
			step[to] = step[from] + 1;
		}
		to = from * 2;
		if (to < N && step[to] == -1)
		{
			q.push(to);
			step[to] = step[from] + 1;
		}
	}
	return step[k];
}
int main()
{
	int n, k;
	while (~scanf("%d%d", &n, &k))
	{
		memset(step, -1, sizeof(step));
		int ans = bfs(n, k);
		printf("%d\n", ans);
	}
	return 0;
}