Question

我有一个100,100个瓷砖的空白网格。起点是（0,0），目标是（99,99）。瓷砖是4路连接。

我的Floodfill算法在30ms内找到最短路径，但我的A *实现速度慢了大约10倍。

注意：无论网格或布局的大小如何，A *都会比我的填充更慢（3 - 10x）。因为洪水填充很简单，所以我怀疑我在A *中缺少某种优化。

这是功能。我使用Python的heapq来维护一个f排序列表。 “图表”包含所有节点，目标，邻居和g / f值。

import heapq

def solve_astar(graph):

    open_q = []

    heapq.heappush(open_q, (0, graph.start_point))

    while open_q:

        current = heapq.heappop(open_q)[1]

        current.seen = True # Equivalent of being in a closed queue

        for n in current.neighbours:
            if n is graph.end_point:
                n.parent = current
                open_q = [] # Clearing the queue stops the process

            # Ignore if previously seen (ie, in the closed queue)
            if n.seen:
                continue

            # Ignore If n already has a parent and the parent is closer
            if n.parent and n.parent.g <= current.g:
                continue

            # Set the parent, or switch parents if it already has one
            if not n.parent:
                n.parent = current
            elif n.parent.g > current.g:
                remove_from_heap(n, n.f, open_q)
                n.parent = current

            # Set the F score (simple, uses Manhattan)
            set_f(n, n.parent, graph.end_point)

            # Push it to queue, prioritised by F score
            heapq.heappush(open_q, (n.f, n))

def set_f(point, parent, goal):
    point.g += parent.g
    h = get_manhattan(point, goal)
    point.f = point.g + h

Answer 1

这是一个打破平局的问题。在空网格上，从（0,0）开始到（99,99）会生成许多具有相同f分数的图块。

通过在启发式中添加微小的微调，将首先选择稍微靠近目的地的图块，这意味着可以更快地达到目标并且需要检查更少的图块。

def set_f(point, parent, goal):
    point.g += parent.g
    h = get_manhattan(point, goal) * 1.001
    point.f = point.g + h

这导致了大约100倍的改进，使其比洪水填充更快。

为什么我的A *实施比洪水填充慢？

1 个答案: