Union-Find 算法（中文稱並查集算法）是解決動態連通性（Dynamic Conectivity）問題的一種算法，作者以此為實例，講述了如何分析和改進算法，本節涉及三個算法實現，分別是Quick Find, Quick Union 和 Weighted Quick Union。
動態連通性(Dynamic Connectivity）
動態連通性是計算機圖論中的一種數據結構，動態維護圖結構中相連接的組信息。
簡單的說就是，圖中各個點之間是否相連、連接后組成了多少個組等信息。我們稱連接在一起就像形成了一個圈子似的，成為一個組(Component)，每個組有其自己的一些特征，比如組內所有成員都有同一個標記等。
提到圈子，大家比較好理解，我們在社交網絡中，彼此熟悉的人之間組成自己的圈子，"熟悉"用計算機中的語言來表示就是“Connected 連通的。圈子是會變化的，今天你又新認識了某人，明天你跟某人友盡了，這種變化是動態的，所以有了動態連通性這種數據結構和問題。
比較常見的應用有，社交網絡中比如LinkedIn, 判斷某個用戶與其它用戶是否熟悉：如果你與用戶A熟悉，用戶A與用戶B熟悉，則認為你與用戶B也是連接的，你可以看到用戶B的信息。在計算機網絡中，也存在類似的情況，判斷網絡中某兩個節點是否相連。
Dynamic Connectivity 的計算機語言表述
給定一個整數對(p,q)，如果p 和 q尚未連通，則使二者相連通。p 和 q 相連后，我們稱 p 和 q 在同一個組內。
當p 和 q 連通時，以下關系則成立：
- 自反性：p和p自身是相連的
- 對稱性：如果p和q相連，那么q和p也相連
- 傳遞性：如果p和q相連，q和r相連，那么p和r也相連
在一個網絡中，會存在很多類似的整數對(p,q)，假設網絡容量是 N，我們可以定義一個從 0 到 N-1的整數數組，p,q是其中的值，我們可能需要的操作有：
- 判斷 p 和 q 是否相連
- 如果未相連，則連接 p 和 q, 如果已相連，則可以不做啥
- 查找 p 或 q 屬於哪個組中 (如圈子）
這里的一個關鍵是，如何確定 p 和 q 是在同一個組內。這意味着，每個組需要有一些特定的屬性，我們在后面的算法中會有考慮。
Union-Find 算法描述 Dynamic Connectivity
Union-Find 算法中，提供了對應的方法來實現我們前面提到的可能的操作：
-
connected(): 判斷 p 和 q 是否相連，這里要調用 find(p) 和 find(q)，如果二者屬於同一個組，則認為是相連的，即isConnected()返回true.
-
union(): 如果未相連，則連接 p 和 q, 如果已相連，則可以不做啥
-
find(): 查找 p 或 q 屬於哪個組中 (如圈子），這里返回值是整數，作為組的標識符(component identifier)。
-
count(): 返回組的數量
算法4中的API:
class UF:
def __init__(self,N):
def union(self,p,q): # initialize N sites with integer names
def find(self,p): #return component identifier for p
def connected(self,p,q): #return true if p and q are in the same component
def count(): #number of components
Union-Find 算法及實現
根據我們前面的描述，如果確定每個組的標識符似乎比較關鍵，只要確定了，就可以判斷是否相連。
那用什么來作為標識符，區分各個組呢？
最簡單的一個辦法是，所有的節點都賦予一個 ID，如果兩個節點相連，則將這兩個節點的 ID 設成一樣的，這樣，這兩個節點便屬於同一個組了。網絡中每個組都有了一個唯一的 ID。只要節點 p 和 q 的 ID 相同，則認為節點 p 和 q 相連。我們用數組來放置節點 ID，find()方法可以快速返回 ID，所以我們的第一個算法就叫做 QuickFind。
QuickFind 算法
QuickFind 算法中，find方法比較簡單，union(p,q)方法需要考慮的一點是，要將與p相連的所有節點 id 都設為q當前的 id，使p所在的組和q所在的組結合成了一個同一組。（注：也可以把與q相連的所有節點id都設為p的id）
最開始的時候，所有節點都互不相連。我們假設所有的節點由id=0到N-1的整數表示。
代碼：
# -*- coding: utf-8 -*-
class QuickFind(object):
id=[]
count=0
def __init__(self,n):
self.count = n
i=0
while i<n:
self.id.append(i)
i+=1
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):
idp = self.find(p)
if not self.connected(p,q):
for i in range(len(self.id)):
if self.id[i]==idp: # 將p所在組內的所有節點的id都設為q的當前id
self.id[i] = self.id[q]
self.count -= 1
我們的測試端代碼如下：
# -*- coding: utf-8 -*-
import quickfind
qf = quickfind.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 k in list:
p = k[0]
q = k[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 components is: %d" % qf.count
運行結果：
initial id list is 0,1,2,3,4,5,6,7,8,9
4 and 3 is connected? True
3 and 8 is connected? True
6 and 5 is connected? True
9 and 4 is connected? True
2 and 1 is connected? True
8 and 9 is connected? True
5 and 0 is connected? True
7 and 2 is connected? True
6 and 1 is connected? True
1 and 0 is connected? True
6 and 7 is connected? True
final id list is 1,1,1,8,8,1,1,1,8,8
count of components is: 2
下圖是算法4中的圖示，可供參考：
QuickFind 算法分析：
find方法快速返回數組的值，但union方法最壞情況下，幾乎需要遍歷整個數組，如果數組很大（比如社交網絡巨大） 、需要連接的節點對很多的時候，QuickFind算法的復雜度就相當大了。所以我們需要改進一下union方法。
QuickUnion 算法
前面的QuickFind算法中，union的時候可能需要遍歷整個數組，導致算法性能下降。有沒有什么辦法可以不用遍歷整個數組，又可以保證同一個組內的所有節點都有一個共同屬性呢？樹結構。樹的所有節點都有一個共同的根節點，每個樹只有一個根節點，那每個樹就可以代表一個組。union(p,q)的時候，只要把p所在的樹附加到q所在的樹的根節點，這樣，p和q就在同一樹中了。
改進后的算法即是QuickUnion算法。我們同樣要用到 id 數組，只是這里的 id 放的是節點所在樹的根節點。
find(p): 返回的是 p 所在樹的根節點
union(p,q): 將 p 所在樹的根節點的 id 設為 q 所在樹的根節點
代碼實現：
# -*- coding: utf-8 -*-
class QuickUnion(object):
id=[]
count=0
def __init__(self,n):
self.count = n
i=0
while i<n:
self.id.append(i)
i+=1
def connected(self,p,q):
if self.find(p) == self.find(q):
return True
else:
return False
def find(self,p):
while (p != self.id[p]):
p = self.id[p]
return p
def union(self,p,q):
idq = self.find(q)
idp = self.find(p)
if not self.connected(p,q):
self.id[idp]=idq
self.count -=1
類似的測試端代碼：
# -*- coding: utf-8 -*-
import quickunion
qf = quickunion.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 k in list:
p = k[0]
q = k[1]
qf.union(p,q)
print "%d and %d is connected? %s" % (p,q,str(qf.connected(p,q) ))
print "final root list is %s" % (",").join(str(x) for x in qf.id)
print "count of components is: %d" % qf.count
運行結果：
initial id list is 0,1,2,3,4,5,6,7,8,9
4 and 3 is connected? True
3 and 8 is connected? True
6 and 5 is connected? True
9 and 4 is connected? True
2 and 1 is connected? True
8 and 9 is connected? True
5 and 0 is connected? True
7 and 2 is connected? True
6 and 1 is connected? True
1 and 0 is connected? True
6 and 7 is connected? True
final root list is 1,1,1,8,3,0,5,1,8,8
count of components is: 2
算法4中的圖示供參考理解：
QuickUnion 算法分析：
union方法已經很快速了現在，find方法比QuickFind慢了，其最壞的情況下，如下圖，一次find需要訪問1+..+N次數組，union方法中需要調用兩次find方法，即復雜度變成2(1+...+N)=(N+1)N，接近N的平方了。
Weighted Quick Union 算法
前面的QuickUnion算法中，union的時候只是簡單的將兩個樹合並起來，並沒有考慮兩個樹的大小，所以導致最壞情況的發生。改進的方法可以是，在union之前，先判斷兩個樹的大小（節點數量），將小點的樹附加到大點的樹上，這樣，合並后的樹的深度不會變得非常大。
示例如下：
要判斷樹的大小，需要引進一個新的數組，size 數組，存放樹的大小。初始化的時候 size 各元素都設為 1。
代碼：
# -*- coding: utf-8 -*-
class WeightedQuickUnion(object):
id=[]
count=0
sz=[]
def __init__(self,n):
self.count = n
i=0
while i<n:
self.id.append(i)
self.sz.append(1) # inital size of each tree is 1
i+=1
def connected(self,p,q):
if self.find(p) == self.find(q):
return True
else:
return False
def find(self,p):
while (p != self.id[p]):
p = self.id[p]
return p
def union(self,p,q):
idp = self.find(p)
print "id of %d is: %d" % (p,idp)
idq = self.find(q)
print "id of %d is: %d" % (q,idq)
if not self.connected(p,q):
print "Before Connected: tree size of %d's id is: %d" % (p,self.sz[idp])
print "Before Connected: tree size of %d's id is: %d" % (q,self.sz[idq])
if (self.sz[idp] < self.sz[idq]):
print "tree size of %d's id is smaller than %d's id" %(p,q)
print "id of %d's id (%d) is set to %d" % (p,idp,idq)
self.id[idp] = idq
print "tree size of %d's id is incremented by tree size of %d's id" %(q,p)
self.sz[idq] += self.sz[idp]
print "After Connected: tree size of %d's id is: %d" % (p,self.sz[idp])
print "After Connected: tree size of %d's id is: %d" % (q,self.sz[idq])
else:
print "tree size of %d's id is larger than or equal with %d's id" %(p,q)
print "id of %d's id (%d) is set to %d" % (q,idq,idp)
self.id[idq] = idp
print "tree size of %d's id is incremented by tree size of %d's id" %(p,q)
self.sz[idp] += self.sz[idq]
print "After Connected: tree size of %d's id is: %d" % (p,self.sz[idp])
print "After Connected: tree size of %d's id is: %d" % (q,self.sz[idq])
self.count -=1
測試端代碼：
# -*- coding: utf-8 -*-
import weightedquickunion
qf = weightedquickunion.WeightedQuickUnion(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 k in list:
p = k[0]
q = k[1]
print "." * 10 + "unioning %d and %d" % (p,q) + "." * 10
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 components is: %d" % qf.count
代碼運行結果：
initial id list is 0,1,2,3,4,5,6,7,8,9
..........unioning 4 and 3..........
id of 4 is: 4
id of 3 is: 3
Before Connected: tree size of 4's id is: 1
Before Connected: tree size of 3's id is: 1
tree size of 4's id is larger than or equal with 3's id
id of 3's id (3) is set to 4
tree size of 4's id is incremented by tree size of 3's id
After Connected: tree size of 4's id is: 2
After Connected: tree size of 3's id is: 1
4 and 3 is connected? True
..........unioning 3 and 8..........
id of 3 is: 4
id of 8 is: 8
Before Connected: tree size of 3's id is: 2
Before Connected: tree size of 8's id is: 1
tree size of 3's id is larger than or equal with 8's id
id of 8's id (8) is set to 4
tree size of 3's id is incremented by tree size of 8's id
After Connected: tree size of 3's id is: 3
After Connected: tree size of 8's id is: 1
3 and 8 is connected? True
..........unioning 6 and 5..........
id of 6 is: 6
id of 5 is: 5
Before Connected: tree size of 6's id is: 1
Before Connected: tree size of 5's id is: 1
tree size of 6's id is larger than or equal with 5's id
id of 5's id (5) is set to 6
tree size of 6's id is incremented by tree size of 5's id
After Connected: tree size of 6's id is: 2
After Connected: tree size of 5's id is: 1
6 and 5 is connected? True
..........unioning 9 and 4..........
id of 9 is: 9
id of 4 is: 4
Before Connected: tree size of 9's id is: 1
Before Connected: tree size of 4's id is: 3
tree size of 9's id is smaller than 4's id
id of 9's id (9) is set to 4
tree size of 4's id is incremented by tree size of 9's id
After Connected: tree size of 9's id is: 1
After Connected: tree size of 4's id is: 4
9 and 4 is connected? True
..........unioning 2 and 1..........
id of 2 is: 2
id of 1 is: 1
Before Connected: tree size of 2's id is: 1
Before Connected: tree size of 1's id is: 1
tree size of 2's id is larger than or equal with 1's id
id of 1's id (1) is set to 2
tree size of 2's id is incremented by tree size of 1's id
After Connected: tree size of 2's id is: 2
After Connected: tree size of 1's id is: 1
2 and 1 is connected? True
..........unioning 8 and 9..........
id of 8 is: 4
id of 9 is: 4
8 and 9 is connected? True
..........unioning 5 and 0..........
id of 5 is: 6
id of 0 is: 0
Before Connected: tree size of 5's id is: 2
Before Connected: tree size of 0's id is: 1
tree size of 5's id is larger than or equal with 0's id
id of 0's id (0) is set to 6
tree size of 5's id is incremented by tree size of 0's id
After Connected: tree size of 5's id is: 3
After Connected: tree size of 0's id is: 1
5 and 0 is connected? True
..........unioning 7 and 2..........
id of 7 is: 7
id of 2 is: 2
Before Connected: tree size of 7's id is: 1
Before Connected: tree size of 2's id is: 2
tree size of 7's id is smaller than 2's id
id of 7's id (7) is set to 2
tree size of 2's id is incremented by tree size of 7's id
After Connected: tree size of 7's id is: 1
After Connected: tree size of 2's id is: 3
7 and 2 is connected? True
..........unioning 6 and 1..........
id of 6 is: 6
id of 1 is: 2
Before Connected: tree size of 6's id is: 3
Before Connected: tree size of 1's id is: 3
tree size of 6's id is larger than or equal with 1's id
id of 1's id (2) is set to 6
tree size of 6's id is incremented by tree size of 1's id
After Connected: tree size of 6's id is: 6
After Connected: tree size of 1's id is: 3
6 and 1 is connected? True
..........unioning 1 and 0..........
id of 1 is: 6
id of 0 is: 6
1 and 0 is connected? True
..........unioning 6 and 7..........
id of 6 is: 6
id of 7 is: 6
6 and 7 is connected? True
final id list is 6,2,6,4,4,6,6,2,4,4
count of components is: 2
算法4中的圖示：
