Saturday, October 8, 2016

Find all reachable nodes from every node present in a given set - GeeksforGeeks


Find all reachable nodes from every node present in a given set - GeeksforGeeks
Given an undirected graph and a set of vertices, find all reachable nodes from every vertex present in the given set.
Consider below undirected graph with 2 disconnected components.
GraphEx1
arr[] = {1 , 2 , 5}
Reachable nodes from 1 are  1, 2, 3, 4
Reachable nodes from 2 are 1, 2, 3, 4
Reachable nodes from 5 are 5, 6, 7
Since the given graph is undirected, all vertices that belong to same component have same set of reachable nodes. So we keep track of vertex and component mappings. Every component in the graph is assigned a number and every vertex in this component is assigned this number. We use the visit array for this purpose, the array which is used to keep track of visited vertices in BFS.
For a node u, 
if visit[u] is 0 then
    u has not been visited before
else // if not zero then
   visit[u] represents the component number. 

For any two nodes u and v belonging to same 
component, visit[u] is equal to visit[v]
To store the reachable nodes, use a map m with key as component number and value as a vector which stores all the reachable nodes.
To find reachable nodes for a node u return m[visit[u]]
Look at the pseudo code below in order to understand how to assign component numbers.
componentNum = 0
for i=1 to n 
    If visit[i] is NOT 0 then
        componentNum++ 
         
        // bfs() returns a list (or vector)
        // for given vertex 'i'
        list = bfs(i, componentNum)
        m[visit[i]]] = list
For the graph shown in the example the visit array would be.
VisitArray (2)
For the nodes 1, 2, 3 and 4 the component number is 1. For nodes 5, 6 and 7 the component number is 2.
Time Complexity Analysis:
n = Size of the given set
E = Number of Edges
V = Number of Nodes
O(V+E) for BFS
In worst case all the V nodes are displayed for each node present in the given, i.e only one component in the graph so it takes O(n*V) time.
Worst Case Time Complexity : O(V+E) + O(n*V)
class Graph
{
public:
    int V;    // No. of vertices
 
    // Pointer to an array containing adjacency lists
    list<int> *adj;
 
    Graph(int );  // Constructor
 
    void addEdge(int, int);
 
    vector<int> BFS(int, int, int []);
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V+1];
}
 
void Graph::addEdge(int u, int v)
{
    adj[u].push_back(v); // Add w to v’s list.
    adj[v].push_back(u); // Add v to w’s list.
}
 
vector<int> Graph::BFS(int componentNum, int src,
                                    int visited[])
{
    // Mark all the vertices as not visited
    // Create a queue for BFS
    queue<int> queue;
 
    queue.push(src);
 
    // Assign Component Number
    visited[src] = componentNum;
 
    // Vector to store all the reachable nodes from 'src'
    vector<int> reachableNodes;
 
    while(!queue.empty())
    {
        // Dequeue a vertex from queue
        int u = queue.front();
        queue.pop();
 
        reachableNodes.push_back(u);
 
        // Get all adjacent vertices of the dequeued
        // vertex u. If a adjacent has not been visited,
        // then mark it visited nd enqueue it
        for (auto itr = adj[u].begin();
                itr != adj[u].end(); itr++)
        {
            if (!visited[*itr])
            {
                // Assign Component Number to all the
                // reachable nodes
                visited[*itr] = componentNum;
                queue.push(*itr);
            }
        }
    }
    return reachableNodes;
}
 
// Display all the Reachable Nodes from a node 'n'
void displayReachableNodes(int n,
            unordered_map <int, vector<int> > m)
{
    vector<int> temp = m[n];
    for (int i=0; i<temp.size(); i++)
        cout << temp[i] << " ";
 
    cout << endl;
}
 
// Find all the reachable nodes for every element
// in the arr
void findReachableNodes(Graph g, int arr[], int n)
{
    // Get the number of nodes in the graph
    int V = g.V;
 
    // Take a integer visited array and initialize
    // all the elements with 0
    int visited[V+1];
    memset(visited, 0, sizeof(visited));
 
    // Map to store list of reachable Nodes for a
    // given node.
    unordered_map <int, vector<int> > m;
 
    // Initialize component Number with 0
    int componentNum = 0;
 
    // For each node in arr[] find reachable
    // Nodes
    for (int i = 0 ; i < n ; i++)
    {
        int u = arr[i];
 
        // Visit all the nodes of the component
        if (!visited[u])
        {
            componentNum++;
 
            // Store the reachable Nodes corresponding to
            // the node 'i'
            m[visited[u]] = g.BFS(componentNum, u, visited);
        }
 
        // At this point, we have all reachable nodes
        // from u, print them by doing a look up in map m.
        cout << "Reachable Nodes from " << u <<" are\n";
        displayReachableNodes(visited[u], m);
    }
}

One straight forward solution is to do a BFS traversal for every node present in the set and then find all the reachable nodes.
Assume that we need to find reachable nodes for n nodes, the time complexity for this solution would be O(n*(V+E)) where V is number of nodes in the graph and E is number of edges in the graph. Please note that we need to call BFS as a separate call for every node without using the visited array of previous traversals because a same vertex may need to be printed multiple times. This seems to be an effective solution but consider the case when E = Θ(V2) and n = V, time complexity becomes O(V3).
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