Weighted Quick Union即:
在Quick Union的基础上对结点加权(weighted),在parent[i]基础上增加一个size[i].
用来存储该结点(site)的所有子结点数目.(size[i] == number of sites in subtree rooted at i)
具体操作步骤:
仅仅在union() operation改变,在改变parent前,增加一个步骤:
比较两个结点的size,谁更轻谁就在下面,
具体看代码:
class WeightedQuickUnion():
__count = int() #number of components
__parent = list() #__parent[i] parent of i
__size = list() #size[i] number of sites in subtree rooted at i
#Each site is initially in its own component
def __init__(self,N):
self.__count = N
for i in range(0,self.__count):
self.__parent.append(i)
self.__size.append(1)
#Return the component identifier for the component containing site
def find(self,p):
self.validate(p)
while (p != self.__parent[p]):
p = self.__parent[p]
return p
def connected(self,p,q):
return self.find(p) == self.find(q)
#Merges the component containig site p with
#the component containing site q
def union(self,p,q):
rootP=self.find(p)
rootQ=self.find(q)
if (rootP == rootQ):
return
if (self.__size[rootP] < self.__size[rootQ]):
self.__parent[rootP] = rootQ
self.__size[rootQ] += self.__size[rootP]
else:
self.__parent[rootQ] = rootP
self.__size[rootP] += self.__size[rootQ]
self.__count-=1
def validate(self, p):
n = len(self.__parent)
if (p < 0 or p >= n):
raise ValueError("index", p, "is not between 0 and", (n - 1))
def traversal(self):
for i in self.__parent:
print(i,end=' ')
WQU = WeightedQuickUnion(8)
WQU.union(0,1)
WQU.union(2,1)
WQU.union(2,4)
WQU.union(3,7)
print(WQU.connected(0,4))
WQU.traversal()
实例同上一文Quick Find一样,
连接0-1-2-4 3-7,并调用connected()方法验证0-4是否连接
最后遍历
输出:
True
0 0 0 3 0 5 6 3
根据输出可以反应出树形图:
0 3 5 6
/ | \ |
1 2 4 7
程序中:union()方法里,把size的比较分为两种情况,小于以及大于+等于.
union(0,1)的时候0是p,1是q,他们的size都是1,所以会执行
self.__parent[rootQ] = rootP
self.__size[rootP] += self.__size[rootQ]
也就是q(1)会成为p(0)的子节点.
union(3,7)同理.
其它情况由于size不同,会按照程序写的那样,把轻的作重的子节点