C++ program to find if a given undirected graph is biconnected

Code:

#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 *adj;    // A dynamic array of adjacency lists
    bool isBCUtil(int v, bool visited[], int disc[], int low[],
                 int parent[]);
public:
    Graph(int V);   // Constructor
    void addEdge(int v, int w); // to add an edge to graph
    bool isBC();    // returns true if graph is Biconnected
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list[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 returns true if there is an articulation
// point in given graph, otherwise returns false.
// This function is almost same as isAPUtil() here ( http://goo.gl/Me9Fw )
// 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
bool Graph::isBCUtil(int u, bool visited[], int disc[],int low[],int parent[])
{
    // 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::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;

            // check if subgraph rooted with v has an articulation point
            if (isBCUtil(v, visited, disc, low, parent))
               return true;

            // 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)
               return 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])
               return true;
        }

        // Update low value of u for parent function calls.
        else if (v != parent[u])
            low[u]  = min(low[u], disc[v]);
    }
    return false;
}

// The main function that returns true if graph is Biconnected, 
// otherwise false. It uses recursive function isBCUtil()
bool Graph::isBC()
{
    // 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];

    // Initialize parent and visited, and ap(articulation point) 
    //  arrays
    for (int i = 0; i < V; i++)
    {
        parent[i] = NIL;
        visited[i] = false;
    }

    // Call the recursive helper function to find if there is an articulation 
    // point in given graph. We do DFS traversal starring from vertex 0
    if (isBCUtil(0, visited, disc, low, parent) == true)
        return false;

    // Now check whether the given graph is connected or not. An undirected
    // graph is connected if all vertices are reachable from any starting 
    // point (we have taken 0 as starting point)
    for (int i = 0; i < V; i++)
        if (visited[i] == false)
            return false;

    return true;
}

// Driver program to test above function
int main()
{
    // Create graphs given in above diagrams
    Graph g1(2);
    g1.addEdge(0, 1);
    g1.isBC()? cout << "Yes\n" : cout << "No\n";

    Graph g2(5);
    g2.addEdge(1, 0);
    g2.addEdge(0, 2);
    g2.addEdge(2, 1);
    g2.addEdge(0, 3);
    g2.addEdge(3, 4);
    g2.addEdge(2, 4);
    g2.isBC()? cout << "Yes\n" : cout << "No\n";

    Graph g3(3);
    g3.addEdge(0, 1);
    g3.addEdge(1, 2);
    g3.isBC()? cout << "Yes\n" : cout << "No\n";

    Graph g4(5);
    g4.addEdge(1, 0);
    g4.addEdge(0, 2);
    g4.addEdge(2, 1);
    g4.addEdge(0, 3);
    g4.addEdge(3, 4);
    g4.isBC()? cout << "Yes\n" : cout << "No\n";

    Graph g5(3);
    g5.addEdge(0, 1);
    g5.addEdge(1, 2);
    g5.addEdge(2, 0);
    g5.isBC()? cout << "Yes\n" : cout << "No\n";

    return 0;
}



Output:

Yes
Yes
No
No
Yes


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