python代码实现dijkstra算法

求解从1到6的最短路径。

python代码实现：（以A-F代表1-6）

# Dijkstra算法需要三张散列表和一个存储列表用于记录处理过的节点，如下：
processed = []

def build_graph():
    """建立图关系的散列表"""
    graph = {}
    graph["A"] = {}
    graph["A"]["B"] = 1
    graph["A"]["C"] = 12
    graph["B"] = {}
    graph["B"]["C"] = 9
    graph["B"]["D"] = 3
    graph["C"] = {}
    graph["C"]["E"] = 5
    graph["D"] = {}
    graph["D"]["E"] = 13
    graph["D"]["F"] = 15
    graph["D"]["C"] = 4
    graph["E"] = {}
    graph["E"]["F"] = 4
    graph["F"] = {}
    return graph

def build_costs():
    """建立开销的散列表"""
    infinity = float("inf")    # 无穷大
    costs = {}
    costs["B"] = 1
    costs["C"] = 12
    costs["D"] = infinity
    costs["E"] = infinity
    costs["F"] = infinity
    return costs

def build_parents():
    """建立存储父节点的散列表"""
    parents = {}
    parents["B"] = "A"
    parents["C"] = "A"
    parents["D"] = None
    parents["E"] = None
    parents["F"] = None
    return parents

# 下面是正式的算法步骤
def find_low_cost_node(costs):
    """首先要在未处理的节点找出开销最小的节点,在（开销表）中找,如果没有或者全部都已经处理过了，返回初始值None"""
    lowest_cost = float("inf")
    lowest_cost_node = None
    for node,cost in costs.items():
        if cost < lowest_cost and node not in processed:
            lowest_cost = cost
            lowest_cost_node = node
    return lowest_cost_node

def solve_shortest_route(graph,costs,parents):
    """dijkstra算法求解"""
    node = find_low_cost_node(costs)
    while node is not None:
        cost = costs[node]
        neighbors = graph[node]
        for n in neighbors.keys():
            new_cost = cost+neighbors[n]
            if costs[n] > new_cost:
                # 更新开销表
                costs[n] = new_cost
                # 更新父节点表
                parents[n] = node
        processed.append(node)
        node = find_low_cost_node(costs)

    # 输出结果：
    print("graph:",graph)
    print("costs:",costs)
    print("parents:",parents)

if __name__ == '__main__':
    solve_shortest_route(build_graph(),build_costs(),build_parents())