Description

Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 160241   Accepted: 48852

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.

Solutions

1. BFS

def minStep_bfs(start:int, end: int) -> int:
    queue = collections.deque()
    steps = [None] * (end + 1)
    queue.append((start, 0))
    while queue:
        pos, cnt = queue.popleft()
        if 0 <= pos <= end and not steps[pos]:
            steps[pos] = cnt
            if pos == end:
                return steps[pos]
            queue.append((pos - 1, cnt + 1))
            queue.append((pos + 1, cnt + 1))
            queue.append((pos * 2, cnt + 1))

2. DFS

def minStep_dfs(start:int, end: int) -> int:
    # backward thinking
    if start >= end: # only -1
        return start - end
    if end % 2 == 0:
        if start < end // 2:
            return minStep_dfs(start, end // 2) + 1
        elif start == end // 2:
            return 1
        else: # start > end // 2
            return min(end - start, 1 + start - end // 2)   # 1 is one operation of //, start - end // 2 is backward
    else:
        return min(minStep_dfs(start, end + 1), minStep_dfs(start, end - 1)) + 1

　　更精简化的：

def minStep_dfs1(start:int, end: int) -> int:
    # backward thinking
    if start >= end: # only -1
        return start - end
    if end % 2 == 0:
        return min(end - start, minStep_dfs1(start, end // 2) + 1)
    else:
        return min(minStep_dfs1(start, end + 1), minStep_dfs1(start, end - 1)) + 1

def minStep_dfs2(start:int, end: int) -> int:
    if start >= end: # only -1
        return start - end
    if end % 2 == 0:
        return min(end - start, minStep_dfs2(start * 2, end) + 1)
    else:
        return min(minStep_dfs2(start, end + 1), minStep_dfs2(start, end - 1)) + 1

References

  1. bfs+dfs分析—-poj 3278 Catch That Cow
  2. DFS