注：本文翻译自http://www.geeksforgeeks.org/articulation-points-or-cut-vertices-in-a-graph/。如有翻译错误请指正。
一个无向联通图（undirected connected graph）中的顶点，如果去掉它会使图不再联通，就是割点（articulation point/cut vertex）。割点表示一个连通网络的脆弱性——一个点出问题会将整个网络分成2个或更多个部分。在设计可靠的网络中，它们很有用。
对于一个无向联通图，一个割点是一个去掉之后会增加连通分量（connected component）数量的点。
下面是一些例子，割点用红色圆圈圈出：
怎么找到一个图里的全部割点？
一个简单方法就是一个接一个地去掉所有顶点，然后看去掉该点是否导致图不再连通。下面是这种简单方法的步骤：
对于每个顶点v
a) 将v从图中去掉
b) 看图是否仍是连通的（既可以使用BFS，也可以使用DFS）
c) 将v重新加入图
上述方法的时间复杂度为O(V*(V+E))，如果使用邻接表存储图。可以更好吗？
寻找所有割点的O(V+E)算法
方法是使用DFS（depth first search，深度优先搜索）。在DFS中，我们用树的形式跟踪顶点，叫做DFS树。在DFS树中，一个顶点u是另一个顶点v的父节点（parent），如果v已经被u发现（很明显v在图上与u相邻）。在DFS树中，如果以下两个条件之一成立，则u是割点：
1) u是DFS树的根节点，并且有两个以上的子节点。
2) u不是DFS树的根节点，它有一个子节点v，并且以v为根节点的子树中，没有一个顶点有通向任意一个u的祖先的回边。
下面的图片显示了与上面一样的概念，还有一点补充，就是DFS树中的叶节点不可能是割点。
对于给定的图，我们用DFS遍历和附加代码，寻找割点（articulation points，APs）。在DFS遍历中，我们维护一个parent[]数组，parent[u]存储u的父节点。在上面提及的两种情况中，第一种情况很好检测。对于每个顶点，数出其子节点数量。如果目前访问过的顶点u是根节点（parent[u]为NIL），并且有两个以上的子节点，输出。
第二种情况怎么处理？第二种情况更复杂。我们维护一个disc[]数组，存储一个顶点被发现的时刻（译者注：即从根节点开始DFS，第几次才能到达该顶点）。对于每个节点u，我们需要找到最早访问过的顶点（访问时刻最低的顶点），且该顶点可以从以u为根节点的子树（的某个顶点）到达。因此我们维护一个额外的数组low[]，定义如下。
low[u] = min(disc[u], disc[w])
w为u的祖先，并且存在一条回边，可以从u的后代到达w。
下面是用来寻找割点的Tarjan算法的C++、Java和Python实现。
C++
// A C++ program to find articulation points in an undirected graph
#include<iostream>
#include <list>
#define NIL -1
using namespace std;
// A class that represents an undirected graph
class Graph
{
int V; // No. of vertices
list<int> *adj; // A dynamic array of adjacency lists
void APUtil(int v, bool visited[], int disc[], int low[],
int parent[], bool ap[]);
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void AP(); // prints articulation points
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v); // Note: the graph is undirected
}
// A recursive function that find articulation points using DFS traversal
// u --> The vertex to be visited next
// visited[] --> keeps tract of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
// ap[] --> Store articulation points
void Graph::APUtil(int u, bool visited[], int disc[],
int low[], int parent[], bool ap[])
{
// A static variable is used for simplicity, we can avoid use of static
// variable by passing a pointer.
static int time = 0;
// Count of children in DFS Tree
int children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices aadjacent to this
list<int>::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
{
int v = *i; // v is current adjacent of u
// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
APUtil(v, visited, disc, low, parent, ap);
// Check if the subtree rooted with v has a connection to
// one of the ancestors of u
low[u] = min(low[u], low[v]);
// u is an articulation point in following cases
// (1) u is root of DFS tree and has two or more chilren.
if (parent[u] == NIL && children > 1)
ap[u] = true;
// (2) If u is not root and low value of one of its child is more
// than discovery value of u.
if (parent[u] != NIL && low[v] >= disc[u])
ap[u] = true;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = min(low[u], disc[v]);
}
}
// The function to do DFS traversal. It uses recursive function APUtil()
void Graph::AP()
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
int *disc = new int[V];
int *low = new int[V];
int *parent = new int[V];
bool *ap = new bool[V]; // To store articulation points
// Initialize parent and visited, and ap(articulation point) arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
ap[i] = false;
}
// Call the recursive helper function to find articulation points
// in DFS tree rooted with vertex ‘i‘
for (int i = 0; i < V; i++)
if (visited[i] == false)
APUtil(i, visited, disc, low, parent, ap);
// Now ap[] contains articulation points, print them
for (int i = 0; i < V; i++)
if (ap[i] == true)
cout << i << " ";
}
// Driver program to test above function
int main()
{
// Create graphs given in above diagrams
cout << "\nArticulation points in first graph \n";
Graph g1(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.AP();
cout << "\nArticulation points in second graph \n";
Graph g2(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.AP();
cout << "\nArticulation points in third graph \n";
Graph g3(7);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.addEdge(2, 0);
g3.addEdge(1, 3);
g3.addEdge(1, 4);
g3.addEdge(1, 6);
g3.addEdge(3, 5);
g3.addEdge(4, 5);
g3.AP();
return 0;
}
Java
// A Java program to find articulation points in an undirected graph
import java.io.*;
import java.util.*;
import java.util.LinkedList;
// This class represents an undirected graph using adjacency list
// representation
class Graph
{
private int V; // No. of vertices
// Array of lists for Adjacency List Representation
private LinkedList<Integer> adj[];
int time = 0;
static final int NIL = -1;
// Constructor
Graph(int v)
{
V = v;
adj = new LinkedList[v];
for (int i=0; i<v; ++i)
adj[i] = new LinkedList();
}
//Function to add an edge into the graph
void addEdge(int v, int w)
{
adj[v].add(w); // Add w to v‘s list.
adj[w].add(v); //Add v to w‘s list
}
// A recursive function that find articulation points using DFS
// u --> The vertex to be visited next
// visited[] --> keeps tract of visited vertices
// disc[] --> Stores discovery times of visited vertices
// parent[] --> Stores parent vertices in DFS tree
// ap[] --> Store articulation points
void APUtil(int u, boolean visited[], int disc[],
int low[], int parent[], boolean ap[])
{
// Count of children in DFS Tree
int children = 0;
// Mark the current node as visited
visited[u] = true;
// Initialize discovery time and low value
disc[u] = low[u] = ++time;
// Go through all vertices aadjacent to this
Iterator<Integer> i = adj[u].iterator();
while (i.hasNext())
{
int v = i.next(); // v is current adjacent of u
// If v is not visited yet, then make it a child of u
// in DFS tree and recur for it
if (!visited[v])
{
children++;
parent[v] = u;
APUtil(v, visited, disc, low, parent, ap);
// Check if the subtree rooted with v has a connection to
// one of the ancestors of u
low[u] = Math.min(low[u], low[v]);
// u is an articulation point in following cases
// (1) u is root of DFS tree and has two or more chilren.
if (parent[u] == NIL && children > 1)
ap[u] = true;
// (2) If u is not root and low value of one of its child
// is more than discovery value of u.
if (parent[u] != NIL && low[v] >= disc[u])
ap[u] = true;
}
// Update low value of u for parent function calls.
else if (v != parent[u])
low[u] = Math.min(low[u], disc[v]);
}
}
// The function to do DFS traversal. It uses recursive function APUtil()
void AP()
{
// Mark all the vertices as not visited
boolean visited[] = new boolean[V];
int disc[] = new int[V];
int low[] = new int[V];
int parent[] = new int[V];
boolean ap[] = new boolean[V]; // To store articulation points
// Initialize parent and visited, and ap(articulation point)
// arrays
for (int i = 0; i < V; i++)
{
parent[i] = NIL;
visited[i] = false;
ap[i] = false;
}
// Call the recursive helper function to find articulation
// points in DFS tree rooted with vertex ‘i‘
for (int i = 0; i < V; i++)
if (visited[i] == false)
APUtil(i, visited, disc, low, parent, ap);
// Now ap[] contains articulation points, print them
for (int i = 0; i < V; i++)
if (ap[i] == true)
System.out.print(i+" ");
}
// Driver method
public static void main(String args[])
{
// Create graphs given in above diagrams
System.out.println("Articulation points in first graph ");
Graph g1 = new Graph(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.AP();
System.out.println();
System.out.println("Articulation points in Second graph");
Graph g2 = new Graph(4);
g2.addEdge(0, 1);
g2.addEdge(1, 2);
g2.addEdge(2, 3);
g2.AP();
System.out.println();
System.out.println("Articulation points in Third graph ");
Graph g3 = new Graph(7);
g3.addEdge(0, 1);
g3.addEdge(1, 2);
g3.addEdge(2, 0);
g3.addEdge(1, 3);
g3.addEdge(1, 4);
g3.addEdge(1, 6);
g3.addEdge(3, 5);
g3.addEdge(4, 5);
g3.AP();
}
}
// This code is contributed by Aakash Hasija
Python
# Python program to find articulation points in an undirected graph
from collections import defaultdict
#This class represents an undirected graph
#using adjacency list representation
class Graph:
def __init__(self,vertices):
self.V= vertices #No. of vertices
self.graph = defaultdict(list) # default dictionary to store graph
self.Time = 0
# function to add an edge to graph
def addEdge(self,u,v):
self.graph[u].append(v)
self.graph[v].append(u)
‘‘‘A recursive function that find articulation points
using DFS traversal
u --> The vertex to be visited next
visited[] --> keeps tract of visited vertices
disc[] --> Stores discovery times of visited vertices
parent[] --> Stores parent vertices in DFS tree
ap[] --> Store articulation points‘‘‘
def APUtil(self,u, visited, ap, parent, low, disc):
#Count of children in current node
children =0
# Mark the current node as visited and print it
visited[u]= True
# Initialize discovery time and low value
disc[u] = self.Time
low[u] = self.Time
self.Time += 1
#Recur for all the vertices adjacent to this vertex
for v in self.graph[u]:
# If v is not visited yet, then make it a child of u
# in DFS tree and recur for it
if visited[v] == False :
parent[v] = u
children += 1
self.APUtil(v, visited, ap, parent, low, disc)
# Check if the subtree rooted with v has a connection to
# one of the ancestors of u
low[u] = min(low[u], low[v])
# u is an articulation point in following cases
# (1) u is root of DFS tree and has two or more chilren.
if parent[u] == -1 and children > 1:
ap[u] = True
#(2) If u is not root and low value of one of its child is more
# than discovery value of u.
if parent[u] != -1 and low[v] >= disc[u]:
ap[u] = True
# Update low value of u for parent function calls
elif v != parent[u]:
low[u] = min(low[u], disc[v])
#The function to do DFS traversal. It uses recursive APUtil()
def AP(self):
# Mark all the vertices as not visited
# and Initialize parent and visited,
# and ap(articulation point) arrays
visited = [False] * (self.V)
disc = [float("Inf")] * (self.V)
low = [float("Inf")] * (self.V)
parent = [-1] * (self.V)
ap = [False] * (self.V) #To store articulation points
# Call the recursive helper function
# to find articulation points
# in DFS tree rooted with vertex ‘i‘
for i in range(self.V):
if visited[i] == False:
self.APUtil(i, visited, ap, parent, low, disc)
for index, value in enumerate (ap):
if value == True: print index,
# Create a graph given in the above diagram
g1 = Graph(5)
g1.addEdge(1, 0)
g1.addEdge(0, 2)
g1.addEdge(2, 1)
g1.addEdge(0, 3)
g1.addEdge(3, 4)
print "\nArticulation points in first graph "
g1.AP()
g2 = Graph(4)
g2.addEdge(0, 1)
g2.addEdge(1, 2)
g2.addEdge(2, 3)
print "\nArticulation points in second graph "
g2.AP()
g3 = Graph (7)
g3.addEdge(0, 1)
g3.addEdge(1, 2)
g3.addEdge(2, 0)
g3.addEdge(1, 3)
g3.addEdge(1, 4)
g3.addEdge(1, 6)
g3.addEdge(3, 5)
g3.addEdge(4, 5)
print "\nArticulation points in third graph "
g3.AP()
#This code is contributed by Neelam Yadav
时间复杂度：上面的函数是一个含有额外数组的简单DFS。因此时间复杂度与DFS相同，对于用邻接表表示的图，为O(V+E)。
引用
https://www.cs.washington.edu/education/courses/421/04su/slides/artic.pdf
http://www.slideshare.net/TraianRebedea/algorithm-design-and-complexity-course-8
http://faculty.simpson.edu/lydia.sinapova/www/cmsc250/LN250_Weiss/L25-Connectivity.htm
时间： 2024-10-14 19:01:19