问题
无向图最小生成树的Prim算法
思路说明
假设点A,B,C,D,E,F,两点之间有连线的,以及它们的距离分别是:(A-B:7);(A-D:5);(B-C:8);(B-D:9);(B-E:7);(C-E:5);(D-E:15);(D-F:6);(E-F:8);(E-G:9);(F-G:11)
关于Prim算法的计算过程,参与维基百科的词条:普里姆算法
将上述点与点关系以及两点之间距离(边长,有的文献中称之为权重)写成矩阵形式(在list中,每两个点及其之间的距离组成一个tuple)
edges = [ (“A”, “B”, 7),
(“A”, “D”, 5),
(“B”, “C”, 8),
(“B”, “D”, 9),
(“B”, “E”, 7),
(“C”, “E”, 5),
(“D”, “E”, 15),
(“D”, “F”, 6),
(“E”, “F”, 8),
(“E”, “G”, 9),
(“F”, “G”, 11)
]
在下面的解决方法中,要计算出与已经选出的若干个点有相邻关系的点中,相应边长最短的点。这本质上是排序之后取出最小的,因为这种排序是动态的,如果用sorted或者list.sort()之类的方法对list排序,一则速度慢(python中的sort方法对大数据时不是很快),二则代码也长了。幸好python提供了一个非常好用的模块:heapq。这个模块是堆排序方法实现排序,并能够随时取出最小值。简化代码,更重要是提升了速度。
就用这个来解决Prim算法问题了。
解决(Python)
1. #! /usr/bin/env python2. #coding:utf-83. 4. from collections import defaultdict5. from heapq import *6. 7. def prim( vertexs, edges ):8. adjacent_vertex = defaultdict(list)9. 10. """11. 注意:defaultdict(list)必须以list做为变量,可以详细阅读:[collections.defaultdict](https://docs.python.org/2/library/collections.html#collections.defaultdict)12. """13. for v1,v2,length in edges:14. adjacent_vertex[v1].append((length, v1, v2))15. adjacent_vertex[v2].append((length, v2, v1))16. 17. """18. 经过上述操作,将edges列表中各项归类成以某点为dictionary的key,其value则是其相邻的点以及边长。如下:19. 20. defaultdict(<type 'list'>, {'A': [(7, 'A', 'B'), (5, 'A', 'D')],21. 'C': [(8, 'C', 'B'), (5, 'C', 'E')],22. 'B': [(7, 'B', 'A'), (8, 'B', 'C'), (9, 'B', 'D'), (7, 'B', 'E')],23. 'E': [(7, 'E', 'B'), (5, 'E', 'C'), (15, 'E', 'D'), (8, 'E', 'F'), (9, 'E', 'G')],24. 'D': [(5, 'D', 'A'), (9, 'D', 'B'), (15, 'D', 'E'), (6, 'D', 'F')],25. 'G': [(9, 'G', 'E'), (11, 'G', 'F')],26. 'F': [(6, 'F', 'D'), (8, 'F', 'E'), (11, 'F', 'G')]})27. 28. """29. 30. mst = [] #存储最小生成树结果31. 32. chosed = set(vertexs[0])33. 34. """35. vertexs是顶点列表,vertexs = list("ABCDEFG")===>vertexs=['A', 'B', 'C', 'D', 'E', 'F', 'G']36. >> chosed=set(vertexs[0])37. >> chosed38. set(['A'])39. 也就是,首先选一个点(这个点是可以任意选的),以这个点为起点,找其相邻点,以及最短边长。40. 41. """42. 43. #得到adjacent_vertexs_edges中顶点是'A'(nodes[0]='A')的相邻点list,即adjacent_vertexs['A']=[(7,'A','B'),(5,'A','D')]44. 45. adjacent_vertexs_edges = adjacent_vertex[vertexs[0]]46. 47. #将usable_edges加入到堆中,并能够实现用heappop从其中动态取出最小值。关于heapq模块功能,参考python官方文档48. 49. heapify(adjacent_vertexs_edges)50. 51. while adjacent_vertexs_edges:52. #得到某个定点(做为adjacent_vertexs_edges的键)与相邻点距离(相邻点和边长/距离做为该键的值)最小值,并同时从堆中清除。53. w, v1, v2 = heappop(adjacent_vertexs_edges)54. if v2 not in chosed:55. 56. #在used中有第一选定的点'A',上面得到了距离A点最近的点'D',举例是5。将'd'追加到used中57. chosed.add(v2)58. 59. mst.append((v1,v2,w)) #将v1,v2,w,第一次循环就是('A','D',5) append into mst60. 61. #再找与d相邻的点,如果没有在heap中,则应用heappush压入堆内,以加入排序行列62. 63. for next_vertex in adjacent_vertex[v2]:64. if next_vertex[2] not in chosed:65. heappush( adjacent_vertexs_edges,next_vertex)66. return mst67. 68. 69. #test70. vertexs = list("ABCDEFG")71. edges = [ ("A", "B", 7), ("A", "D", 5),72. ("B", "C", 8), ("B", "D", 9),73. ("B", "E", 7), ("C", "E", 5),74. ("D", "E", 15), ("D", "F", 6),75. ("E", "F", 8), ("E", "G", 9),76. ("F", "G", 11)]77. 78. print "edges:",edges79. print "prim:", prim( vertexs, edges )
运行结果
edges: [(‘A’, ‘B’, 7), (‘A’, ‘D’, 5), (‘B’, ‘C’, 8), (‘B’, ‘D’, 9), (‘B’, ‘E’, 7), (‘C’, ‘E’, 5), (‘D’, ‘E’, 15), (‘D’, ‘F’, 6), (‘E’, ‘F’, 8), (‘E’, ‘G’, 9), (‘F’, ‘G’, 11)] prim: [(‘A’, ‘D’, 5), (‘D’, ‘F’, 6), (‘A’, ‘B’, 7), (‘B’, ‘E’, 7), (‘E’, ‘C’, 5), (‘E’, ‘G’, 9)]
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