python实现Floyd算法
(编辑:jimmy 日期: 2024/11/21 浏览:3 次 )
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_list def 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']
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