Kruskal实现最小生成树
算法原理:
一群孤立的顶点,在不形成环的情况下不断把最小的边连接起来。
如何描述不形成环?
1、书上的描述,把连接的顶点,标记为同一个连通分量,线段的起点 和 属于不同的连通分量即可,实现比较简单,只要一个标记数组就可以实现。
2、还有一种描述,不使用连通分量方式,检测回路,类似于 .parent != ,目前思考还不成熟。
代码实现:
def kruskal2(graph):
vnum = len(graph)
pqueue = []
for head in graph:
for tail in graph[head].keys():
heapq.heappush(pqueue,(graph[head][tail],head,tail))
reps = {vertex:vertex for vertex in graph}
mst = {vertex : None for vertex in graph}
count = 0
while count < vnum and pqueue:
pair = heapq.heappop(pqueue)
weight = pair[0]
head = pair[1]
tail = pair[2]
if reps[head] == reps[tail]:
continue
mst[tail]=(head,weight)
for v in graph:
if reps[v] == reps[tail]:
reps[v]=reps[head]
count += 1
return mst
输出结果:
#%%
g = {'A':{'B':1,'C':2},
'B':{'A':1,'C':3,'D':4},
'C':{'A':2,'B':3,'D':5,'E':6},
'D':{'B':4,'C':5,'E':7,'F':8},
'E':{'C':6,'D':7,'G':9},
'F':{'D':8},
'G':{'E':9}
}
t=kruskal(g)
print t
m=kruskal2(g)
print m
{'A': None, 'C': ('A', 2), 'B': ('A', 1), 'E': ('C', 6), 'D': ('B', 4), 'G': ('E', 9), 'F': ('D', 8)}
{'A': None, 'C': ('A', 2), 'B': ('A', 1), 'E': ('C', 6), 'D': ('B', 4), 'G': ('E', 9), 'F': ('D', 8)}