Find the Weak Connected Component in the Directed Graph

Find the number Weak Connected Component in the directed graph. Each node in the graph contains a label and a list of its neighbors. (a connected set of a directed graph is a subgraph in which any two vertices are connected by direct edge path.

Example

Given graph:

A----->B  C
 \     |  | 
  \    |  |
   \   |  |
    \  v  v
     ->D  E <- F

Return {A,B,D}, {C,E,F}. Since there are two connected component which are {A,B,D} and {C,E,F}

这题和Find the Connected Component in the Undirected Graph看起来是很一样的题目，但是实际上在有向图中的弱连通区域是无法走通的。所以使用DFS无法解决这个问题。那么Union Find 并查集就可以出马了，毕竟它不要求边的方向，只要求连接。所以在这题里面，如果一条边的两个结点有未出现的，则先加入点。之后再对每条边的两个结点做按秩合并。所以并查集的实现，这里需要使用hashmap来保存已有结点，实时加入。这题还有一个问题，就是如何输出最后的弱连接块。显然这里的弱连接块，每块的结点的father最终都是同一个，按这个进行查找。代码如下：

# Definition for a directed graph node
# class DirectedGraphNode:
#     def __init__(self, x):
#         self.label = x
#         self.neighbors = []
class Solution:
    # @param {DirectedGraphNode[]} nodes a array of directed graph node
    # @return {int[][]} a connected set of a directed graph
    def connectedSet2(self, nodes):
        hashmap = {}
        UF = UnionFind()
        for node in nodes:
            if node.label not in hashmap:
                hashmap[node.label] = node.label
                UF.add(node.label)
            for neighbor in node.neighbors:
                if neighbor.label not in hashmap:
                    hashmap[neighbor.label] = neighbor.label
                    UF.add(neighbor.label)
                UF.union(node.label, neighbor.label)
        res = {}
        for i in hashmap.keys():
            father = UF.find(i)
            if father not in res:
                res[father] = [i]
            else:
                res[father].append(i)
        ans = map(sorted,res.values())
        return ans

class UnionFind(object):
    def __init__(self):
        self.id = {}
        self.sz = {}
        
    def add(self, x):
        self.id[x] = x
        self.sz[x] = 1
        
    def find(self,x):
        while x != self.id[x]:
            self.id[x] = self.id[self.id[x]]
            x = self.id[x]
        return x
        
    def union(self, x, y):
        i = self.find(x)
        j = self.find(y)
        if i != j:
            if self.sz[i] < self.sz[j]:
                i, j = j, i
            self.id[j] = i #j is added to i
            self.sz[i] += self.sz[j]
            

并查集的结点数目为V，边的数目为E，合并操作一共是E个，所以最终的复杂度是O(V+E）级别的，最终输出结果的复杂度最坏是O(VlogV)级别的，最坏全部在一个弱联通块里面。

转载于:https://www.cnblogs.com/sherylwang/p/5646845.html