Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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.

Source

思路:裸BFS可过
AC代码:
#include <cstdio>
#include <queue>
using namespace std;
queue <int> q;
int n, k;
int step[100010];
bool vis[100010];
void input()
{
	scanf("%d%d", &n, &k);
}
inline bool check(int pos)
{
	if (pos < 0 || pos > 100000 || vis[pos])
		return 0;
	return 1;
}
void bfs()
{
	q.push(n);
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		if (u == k)
		{
			printf("%d\n", step[u]);
			return;
		}
		if (check(u + 1))
		{
			step[u + 1] = step[u] + 1;
			vis[u + 1] = 1;
			q.push(u + 1);
		}
		if (check(u - 1))
		{
			step[u - 1] = step[u] + 1;
			vis[u - 1] = 1;
			q.push(u - 1);
		}
		if (check(u * 2))
		{
			step[u * 2] = step[u] + 1;
			vis[u * 2] = 1;
			q.push(u * 2);
		}
	}
}
int main()
{
	input();
	bfs();
	return 0;
}

 

分类: ACM/ICPC

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