迪克斯特拉 算法(算最短距离)
import time graph = {} # 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 = 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: # 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_cost parents[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)