# -*-coding:utf-8-*-
# 用散列表实现图的关系
# 创建节点的开销表，开销是指从"起点"到该节点的权重
graph = {}
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2

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

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

# 无穷大
infinity = float("inf")
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["end"] = infinity

# 父节点散列表
parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["end"] = None

# 已经处理过的节点，需要记录
processed = []


# 找到开销最小的节点
def find_lowest_cost_node(costs):
 # 初始化数据
 lowest_cost = infinity
 lowest_cost_node = None
 # 遍历所有节点
 for node in costs:
 # 该节点没有被处理
 if not node in processed:
  # 如果当前节点的开销比已经存在的开销小，则更新该节点为开销最小的节点
  if costs[node] < lowest_cost:
  lowest_cost = costs[node]
  lowest_cost_node = node
 return lowest_cost_node


# 找到最短路径
def find_shortest_path():
 node = "end"
 shortest_path = ["end"]
 while parents[node] != "start":
 shortest_path.append(parents[node])
 node = parents[node]
 shortest_path.append("start")
 return shortest_path


# 寻找加权的最短路径
def dijkstra():
 # 查询到目前开销最小的节点
 node = find_lowest_cost_node(costs)
 # 只要有开销最小的节点就循环（这个while循环在所有节点都被处理过后结束）
 while node is not None:
 # 获取该节点当前开销
 cost = costs[node]
 # 获取该节点相邻的节点
 neighbors = graph[node]
 # 遍历当前节点的所有邻居
 for n in neighbors.keys():
  # 计算经过当前节点到达相邻结点的开销,即当前节点的开销加上当前节点到相邻节点的开销
  new_cost = cost + neighbors[n]
  # 如果经当前节点前往该邻居更近，就更新该邻居的开销
  if new_cost < costs[n]:
  costs[n] = new_cost
  #同时将该邻居的父节点设置为当前节点
  parents[n] = node
 # 将当前节点标记为处理过
 processed.append(node)
 # 找出接下来要处理的节点，并循环
 node = find_lowest_cost_node(costs)
 # 循环完毕说明所有节点都已经处理完毕
 shortest_path = find_shortest_path()
 shortest_path.reverse()
 print(shortest_path)
# 测试
dijkstra()