逻辑分析：

1: 找出最便宜的的节点,可在短时间内前往的节点 2: 对于该节点的邻居检查是否有其他节点，如果有就更新开销 3：重复这个过程直到对图中的所有节点都这样做 4：计算最终路径

用于解决的问题

1.从A出发是否存在到达B的路径； 2.从A出发到达B的最短路径(时间最少、或者路径最少等)，事实上最后计算完成后，已经得到了A到各个节点的最短路径了；

#把所有的节点的邻居都存储在grash里面
graph = {}
graph["Start"] = {}
graph["Start"]["A"] = 6
graph["Start"]["B"] = 2
graph["A"] = {}
graph["A"]["End"] = 4
graph["B"] = {}
graph["B"]["C"] = 1
graph["B"]["End"] = 7
graph["C"] = {}
graph["C"]["A"] = 4
graph["C"]["End"] = 5

#infinity代表无穷大（或者未知）
infinity = float("inf")
#costs 代表所有的权重
costs = {}
costs["A"] = 6
costs["B"] = 2
costs["C"] = infinity
costs["End"] = infinity
#用来存储父节点
parents = {}
parents["A"] = "Start"
parents["B"] = "Start"
parents["C"] = None
parents["End"] = None

#用来记录处理过的节点
processed = []

#在未处理的节点里面找出开销最小的节点
def findLowestCostNode(costs, processed):
    lowest_cost = infinity
    lowest_cost_node = None
    for node in costs.keys():
        cost = costs[node]
        if  cost < lowest_cost and node not in processed: #每次找到最小的并且并不存在在poressed里面
            lowest_cost = cost
            lowest_cost_node = node
    return lowest_cost_node
		
def  getResult():
       node = findLowestCostNode(costs, processed) #节点
       while node is not None and node != "End":
        cost = costs[node] # 获取开销
        print(node)
        neighbors = graph[node]
        for n in neighbors.keys():
            new_cost = cost + neighbors[n]
            if new_cost < costs[n]:
                costs[n] = new_cost
                parents[n] = node

        processed.append(node)
        node = findLowestCostNode(costs, processed)
   
if __name__ == '__main__':
      getResult()
      print(costs)
      print(parents)