python实现Floyd算法

Python  /  管理员 发布于 7年前   253

下面是用Python实现Floyd算法的代码，供大家参考，具体内容如下

# -*- coding: utf-8 -*- """ Created on Thu Jul 13 14:56:37 2017  @author: linzr """  ## 表示无穷大 INF_val = 9999  class Floyd_Path():  def __init__(self, node, node_map, path_map):   self.node = node   self.node_map = node_map   self.node_length = len(node_map)   self.path_map = path_map   self._init_Floyd()    def __call__(self, from_node, to_node):   self.from_node = from_node   self.to_node = to_node   return self._format_path()   def _init_Floyd(self):   for k in range(self.node_length):    for i in range(self.node_length):     for j in range(self.node_length):      tmp = self.node_map[i][k] + self.node_map[k][j]      if self.node_map[i][j] > tmp:       self.node_map[i][j] = tmp       self.path_map[i][j] = self.path_map[i][k]          print '_init_Floyd is end'    def _format_path(self):   node_list = []   temp_node = self.from_node   obj_node = self.to_node   print("the shortest path is: %d")%(self.node_map[temp_node][obj_node])   node_list.append(self.node[temp_node])   while True:    node_list.append(self.node[self.path_map[temp_node][obj_node]])    temp_node = self.path_map[temp_node][obj_node]    if temp_node == obj_node:     break;      return node_listdef set_node_map(node_map, node, node_list, path_map):  for i in range(len(node)):   ## 对角线为0   node_map[i][i] = 0  for x, y, val in node_list:   node_map[node.index(x)][node.index(y)] = node_map[node.index(y)][node.index(x)] = val   path_map[node.index(x)][node.index(y)] = node.index(y)   path_map[node.index(y)][node.index(x)] = node.index(x)    if __name__ == "__main__":  node = ['A', 'B', 'C', 'D', 'E', 'F', 'G']  node_list = [('A', 'F', 9), ('A', 'B', 10), ('A', 'G', 15), ('B', 'F', 2),      ('G', 'F', 3), ('G', 'E', 12), ('G', 'C', 10), ('C', 'E', 1),      ('E', 'D', 7)]   ## node_map[i][j] 存储i到j的最短距离  node_map = [[INF_val for val in xrange(len(node))] for val in xrange(len(node))]  ## path_map[i][j]=j 表示i到j的最短路径是经过顶点j  path_map = [[0 for val in xrange(len(node))] for val in xrange(len(node))]    ## set node_map  set_node_map(node_map, node, node_list, path_map)    ## select one node to obj node, e.g. A --> D(node[0] --> node[3])  from_node = node.index('A')  to_node = node.index('E')  Floydpath = Floyd_Path(node, node_map, path_map)  path = Floydpath(from_node, to_node)  print path 

运行结果为：
the shortest path is: 23
['A', 'F', 'G', 'C', 'E']