import timegraph = {}
# 开始节点
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2# a节点
graph["a"] = {}
graph["a"]["fin"] = 1# b节点到下一节点的距离
graph["b"] = {}
graph["b"]["a"] = 3
graph["b"]["fin"] = 5# 结束节点没有邻居节点了
graph["fin"] = {}infinity = float("inf") # 无限大# 创建开销表
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity# 存储到子节点的父节点
parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None# 创造记录过的节点记录表
processed = []def find_lowest_cost_node(costs):lowest_cost = float("inf") # 无穷大lowest_cost_node = None # 最低开销的节点for node in costs: # 遍历所有节点 {"a":6,"b":2,"fin":inf}cost = costs[node]if cost < lowest_cost and node not in processed : # 如果当前节点的开销更低且未处理过，就将其视为开销最低的节点lowest_cost = costlowest_cost_node = nodereturn lowest_cost_nodenode = 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: # 这个while循环在所有节点都被处理过后结束cost = costs[node]neighbors = graph[node]for n in neighbors.keys(): # 遍历当前节点的所有邻居new_cost = cost + neighbors[n]if costs[n] > new_cost: # 如果经当前节点前往该邻居节点更近，就更新邻居节点costs[n] = new_costparents[n] = node # 同时将该邻居的父节点设置为当前节点processed.append(node) # 将当前节点标记为处理过node = find_lowest_cost_node(costs) # 找出接下啦要处理的节点，并循环
print(parents)
print(cost)
