Shortest Path in a weighted Graph where weight of an edge is 1 or 2 - GeeksforGeeks

Shortest Path in a weighted Graph where weight of an edge is 1 or 2 - GeeksforGeeks
Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex 's' to a given destination vertex 't'. Expected time complexity is O(V+E).

A Simple Solution is to use Dijkstra’s shortest path algorithm, we can get a shortest path in O(E + VLogV) time.

How to do it in O(V+E) time? The idea is to use BFS. One important observation about BFS is, the path used in BFS always has least number of edges between any two vertices. So if all edges are of same weight, we can use BFS to find the shortest path. For this problem, we can modify the graph and split all edges of weight 2 into two edges of weight 1 each. In the modified graph, we can use BFS to find the shortest path.

How many new intermediate vertices are needed? We need to add a new intermediate vertex for every source vertex. The reason is simple, if we add a intermediate vertex x between u and v and if we add same vertex between y and z, then new paths u to z and y to v are added to graph which might have note been there in original graph. Therefore in a graph with V vertices, we need V extra vertices.

Below is C++ implementation of above idea. In the below implementation 2*V vertices are created in a graph and for every edge (u, v), we split it into two edges (u, u+V) and (u+V, w). This way we make sure that a different intermediate vertex is added for every source vertex.

How is this approach O(V+E)? In worst case, all edges are of weight 2 and we need to do O(E) operations to split all edges and 2V vertices, so the time complexity becomes O(E) + O(V+E) which is O(V+E).

void Graph::addEdge(int v, int w, int weight)

{

    // split all edges of weight 2 into two

    // edges of weight 1 each.  The intermediate

    // vertex number is maximum vertex number + 1,

    // that is V.

    if (weight==2)

    {

        adj[v].push_back(v+V);

        adj[v+V].push_back(w);

    }

    else // Weight is 1

        adj[v].push_back(w); // Add w to v’s list.

}

 

// To print the shortest path stored in parent[]

int Graph::printShortestPath(int parent[], int s, int d)

{

    static int level = 0;

 

    // If we reached root of shortest path tree

    if (parent[s] == -1)

    {

        cout << "Shortest Path between " << s << " and "

             << d << " is "  << s << " ";

        return level;

    }

 

    printShortestPath(parent, parent[s], d);

 

    level++;

    if (s < V)

        cout << s << " ";

 

    return level;

}

 

// This function mainly does BFS and prints the

// shortest path from src to dest. It is assumed

// that weight of every edge is 1

int Graph::findShortestPath(int src, int dest)

{

    // Mark all the vertices as not visited

    bool *visited = new bool[2*V];

    int *parent = new int[2*V];

 

    // Initialize parent[] and visited[]

    for (int i = 0; i < 2*V; i++)

    {

        visited[i] = false;

        parent[i] = -1;

    }

 

    // Create a queue for BFS

    list<int> queue;

 

    // Mark the current node as visited and enqueue it

    visited[src] = true;

    queue.push_back(src);

 

    // 'i' will be used to get all adjacent vertices of a vertex

    list<int>::iterator i;

 

    while (!queue.empty())

    {

        // Dequeue a vertex from queue and print it

        int s = queue.front();

 

        if (s == dest)

            return printShortestPath(parent, s, dest);

 

        queue.pop_front();

 

        // Get all adjacent vertices of the dequeued vertex s

        // If a adjacent has not been visited, then mark it

        // visited and enqueue it

        for (i = adj[s].begin(); i != adj[s].end(); ++i)

        {

            if (!visited[*i])

            {

                visited[*i] = true;

                queue.push_back(*i);

                parent[*i] = s;

            }

        }

    }

}

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