python实现Dijkstra’s algorithm

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graph = {}  # 存储所有节点的所有邻居和前往邻居的开销
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2

graph["a"] = {}
graph["a"]["fin"] = 1

graph["b"] = {}
graph["b"]["a"] = 3
graph["b"]["fin"] = 5

graph["fin"] = {}

print(graph)  # {'a': {'fin': 1}, 'start': {'a': 6, 'b': 2}, 'b': {'a': 3, 'fin': 5}, 'fin': {}}

infinity = float("inf")  # 无穷大
costs = {}   # 存储前往各个点的开销
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity  #到终点的开销初始值为无限大

parents = {}  #存储各个节点的父节点
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None   #终点的父节点初始值为空

processed = []   # 保存已处理过的节点

#找出开销最小的节点
def find_lowest_cost_node(costs):  
    lowest_cost = float("inf")
    lowest_cost_node = None
    for node in costs:
        cost = costs[node]
        if cost < lowest_cost and node not in processed:
            lowest_cost = cost
            lowest_cost_node = node
    return lowest_cost_node

node = find_lowest_cost_node(costs) #在未处理的节点中找出开销最小的节点
while node is not None:
    cost = costs[node]   # cost = 当前节点的开销
    neighbors = graph[node]   #neighbors = 当前节点的邻居和开销
    for n in neighbors.keys():  #遍历当前节点的所有邻居
        new_cost =  cost + neighbors[n]  # new_cost = 从起点到达邻居n的开销
        if costs[n] > new_cost:   #如果之前记录的邻居n开销大于此次计算的
            costs[n] = new_cost  # 重新设置邻居n的开销
            parents[n] = node   # 设置邻居n的父节点为当前处理的node节点
    
    processed.append(node) #标记为已处理
    node = find_lowest_cost_node(costs) #找出接下来要处理的节点,重新开始循环