在任何一群人中,都有许多对朋友。假设有两个人共享一个朋友,他们自己也是朋友。(是的,这在现实生活中是一个不切实际的假设,但不管怎样,让我们这样做)。换句话说,如果A和B是朋友,B和C是朋友,那么A和C也一定是朋友。使用这个规则,我们可以将任何一组人划分为朋友圈,只要我们对组中的友谊有所了解。
编写一个带有两个参数的函数network()。第一个参数是组中的人数,第二个参数是定义好友的元组对象列表。假设人们是由数字0到n-1标识的。例如,tuple (0,2)表示person 0与person 2是好友。该函数应打印出好友圈子中的人的划分。下面显示了该函数的几个示例运行:
>>>networks(5,[(0,1),(1,2),(3,4)])#execute
社交网络0为{0,1,2}
社交网络1为{3,4}
老实说,我对如何开始这个程序相当迷茫,任何提示都将非常感谢。
发布于 2013-03-21 23:53:06
def networks(n,lst):
groups= []
for i in range(n)
groups.append({i})
for pair in lst:
union = groups[pair[0]]|groups[pair[1]]
for p in union:
groups[p]=union
sets= set()
for g in groups:
sets.add(tuple(g))
i=0
for s in sets:
print("network",i,"is",set(s))
i+=1
如果有人关心的话这就是我要找的。
发布于 2013-03-11 14:44:31
可以用来解决这个问题的一种有效的数据结构是disjoint set
,也称为union-find
结构。不久前,我为another answer写了一个。
结构是这样的:
class UnionFind:
def __init__(self):
self.rank = {}
self.parent = {}
def find(self, element):
if element not in self.parent: # leader elements are not in `parent` dict
return element
leader = self.find(self.parent[element]) # search recursively
self.parent[element] = leader # compress path by saving leader as parent
return leader
def union(self, leader1, leader2):
rank1 = self.rank.get(leader1,1)
rank2 = self.rank.get(leader2,1)
if rank1 > rank2: # union by rank
self.parent[leader2] = leader1
elif rank2 > rank1:
self.parent[leader1] = leader2
else: # ranks are equal
self.parent[leader2] = leader1 # favor leader1 arbitrarily
self.rank[leader1] = rank1+1 # increment rank
下面是你如何使用它来解决你的问题:
def networks(num_people, friends):
# first process the "friends" list to build disjoint sets
network = UnionFind()
for a, b in friends:
network.union(network.find(a), network.find(b))
# now assemble the groups (indexed by an arbitrarily chosen leader)
groups = defaultdict(list)
for person in range(num_people):
groups[network.find(person)].append(person)
# now print out the groups (you can call `set` on `g` if you want brackets)
for i, g in enumerate(groups.values()):
print("Social network {} is {}".format(i, g))
发布于 2013-03-12 12:36:48
这是一个基于connected components in a graph (由@Blckknght推荐)的解决方案:
def make_friends_graph(people, friends):
# graph of friends (adjacency lists representation)
G = {person: [] for person in people} # person -> direct friends list
for a, b in friends:
G[a].append(b) # a is friends with b
G[b].append(a) # b is friends with a
return G
def networks(num_people, friends):
direct_friends = make_friends_graph(range(num_people), friends)
seen = set() # already seen people
# person's friendship circle is a person themselves
# plus friendship circles of all their direct friends
# minus already seen people
def friendship_circle(person): # connected component
seen.add(person)
yield person
for friend in direct_friends[person]:
if friend not in seen:
yield from friendship_circle(friend)
# on Python <3.3
# for indirect_friend in friendship_circle(friend):
# yield indirect_friend
# group people into friendship circles
circles = (friendship_circle(person) for person in range(num_people)
if person not in seen)
# print friendship circles
for i, circle in enumerate(circles):
print("Social network %d is {%s}" % (i, ",".join(map(str, circle))))
示例:
networks(5, [(0,1),(1,2),(3,4)])
# -> Social network 0 is {0,1,2}
# -> Social network 1 is {3,4}
https://stackoverflow.com/questions/15331877
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