Quick Find

# -*- coding: utf-8 -*-
class QuickFind(object):
    id = []
    cnt = 0  # the number of the sets

    def __init__(self, n):
        self.cnt = n
        for i in range(0, n):
            self.id.append(i)

    def connected(self, p, q):
        return self.find(p) == self.find(q)

    def
 
 find(self,p):
        return self.id[p]

    def union(self,p,q):
        id_p = self.find(p)
        if not self.connected(p,q):
            for i in range(0,len(self.id)):
                if self.id[i] == id_p:
                    self.id[i] = self.id[q]  #combine
            self.cnt -= 1

qf = QuickFind(10)

 
print("initial id list is %s" % (‘,‘).join(str(x) for x in qf.id))

list = [
        (4,3),
        (3,8),
        (6,5),
        (9,4),
        (2,1),
        (8,9),
        (5,0),
        (7,2),
        (6,1),
        (1,0),
        (6,7)
        ]

for edge in list :
    p = edge[0]
    q = edge[1]
    qf.union(p,q)
    
 
print("%d and %d is connected? %s"%(p,q,str(qf.connected(p,q))))

print("final id list is %s" %(",").join(str(x) for x in qf.id))
print("count of sets is : %d" % qf.cnt)

connect更新的時候需要遍歷，可以借用樹結構來減小時間復雜度。

# -*- coding: utf-8 -*-
class QuickUnion(object):
    id = []
    cnt = 0  # the number of the sets

    def __init__(self, n):
        self.cnt = n
        for i in range(0, n):
            self.id.append(i)

    def connected(self, p, q):
        return self.find(p) == self.find(q)

    def find(self,x):
        while(x != self.id[x]):
            x = self.id[x]
        return x


    def union(self,p,q):
        root_p = self.find(p)
        root_q = self.find(q)
        if not self.connected(p,q):
            self.id[root_p] = root_q
            self.cnt -= 1

    

qu = QuickUnion(10)

print("initial id list is %s" % (‘,‘).join(str(x) for x in qf.id))

list = [
        (4,3),
        (3,8),
        (6,5),
        (9,4),
        (2,1),
        (8,9),
        (5,0),
        (7,2),
        (6,1),
        (1,0),
        (6,7)
        ]

for edge in list :
    p = edge[0]
    q = edge[1]
    qu.union(p,q)
    print("%d and %d is connected? %s"%(p,q,str(qu.connected(p,q))))

print("final id list is %s" %(",").join(str(x) for x in qu.id))
print("count of sets is : %d" % qu.cnt)

Weighted Quick Union

Quick Union 可能會退化成鏈，導致性能下降，可以通過Weighted Quick Union盡量平衡樹的高度，從而提升性能。

# -*- coding: utf-8 -*-
class WeightedQuickUnion(object):
    id = []
    cnt = 0  # the number of the sets
    sz = []
    def __init__(self, n):
        self.cnt = n
        for i in range(0, n):
            self.id.append(i)
            self.sz.append(1)


    def connected(self, p, q):
        return self.find(p) == self.find(q)

    def find(self,x):
        while(x != self.id[x]):
            x = self.id[x]
        return x


    def union(self,p,q):
        root_p = self.find(p)
        root_q = self.find(q)
        if not self.connected(p,q):
            if self.sz[q] > self.sz[p]:    #保證小樹連大樹，若大樹連小樹可能退化成鏈
                self.id[root_p] = root_q
                self.cnt -= 1
                self.sz[q] += self.sz[p]
            else :
                self.id[root_q] = root_p
                self.cnt -= 1
                self.sz[p] += self.sz[q]




wqu = WeightedQuickUnion(10)

print("initial id list is %s" % (‘,‘).join(str(x) for x in wqu.id))

list = [
        (4,3),
        (3,8),
        (6,5),
        (9,4),
        (2,1),
        (8,9),
        (5,0),
        (7,2),
        (6,1),
        (1,0),
        (6,7)
        ]

for edge in list :
    p = edge[0]
    q = edge[1]
    wqu.union(p,q)
    print("%d and %d is connected? %s"%(p,q,str(wqu.connected(p,q))))

print("final id list is %s" %(",").join(str(x) for x in wqu.id))
print("count of components is : %d" % wqu.cnt)

可通過壓縮查找進一步提升性能，類似樹樁。

參考：http://www.cnblogs.com/learnbydoing/p/6896472.html?utm_source=itdadao&utm_medium=referral

python----並查集