A星算法与Dijkstra算法

Question

我正在接近探路者算法的世界，所以我开始玩两个最着名的A *和Dijkstra。

我正在使用Python开发算法，图形是字典词典：

{Node:{(Node1:weights1),(Node2:weights2),..}

我的空间是640x480的棋盘，这些是可以遵循的所有可能路径：

这是我对算法的实现：

def Manhattan(node1, node2):
    j1 = node1%640
    i1 = (node1-j1)/640
    j2 = node2%640
    i2 = (node2-j2)/640
    return np.abs(j1-j2) + np.abs(i1-i2)

def Astar(graph, start, end):
    A = [None] * (640*480)
    queue = [(0, 0, start, ())]
    while queue:
        fscore, path_len, v, sortPath = heappop(queue)
        sortPath = (v, sortPath)
        if v == end : break
        for w, edge_len in graph[v].items():
            new_cost = path_len+edge_len
            if A[w] is None or new_cost < A[w]:
                A[w] = new_cost
                heappush(queue, (new_cost + Manhattan(w,end), new_cost, w, sortPath))

    # to give same result as original, assign zero distance to unreachable vertices             
    return [0 if x is None else x for x in A],sortPath


def Dijkstra(graph, start, end=None):
    A = [None] * (640*480)
    queue = [(0, start, ())]
    while queue:
        path_len, v, sortPath = heappop(queue)
        sortPath = (v, sortPath)
        if v == end : break
        if A[v] is None: # v is unvisited
            A[v] = path_len
            for w, edge_len in graph[v].items():
                if A[w] is None:
                    heappush(queue, (path_len + edge_len, w, sortPath))

    # to give same result as original, assign zero distance to unreachable vertices             
    return [0 if x is None else x for x in A],sortPath

据我所知，A *算法（如果启发函数有意义;我使用曼哈顿，因为它不可能进入对角线）应该比Dijkstra更快，因为它应该避免扩展不感兴趣的节点。

在我的情况下，Dijkstra在0.28 s中运行而在1 s中运行A *。两种算法的探索节点完全相同：

的Dijkstra：

A *：

我是否对A *实施做错了什么？运行时间的差异，应该是因为启发式函数我每次都从线性索引转换为矩阵行，列？任何帮助是极大的赞赏！