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

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.

C/C++

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// Program to shortest path from a given source vertex ‘s’ to
// a given destination vertex ‘t’.  Expected time complexity
// is O(V+E).
#include<bits/stdc++.h>
using namespace std;
  
// This class represents a directed graph using adjacency
// list representation
class Graph
{
    int V;    // No. of vertices
    list<int> *adj;    // adjacency lists
public:
    Graph(int V);  // Constructor
    void addEdge(int v, int w, int weight); // adds an edge
  
    // finds shortest path from source vertex ‘s’ to
    // destination vertex ‘d’.
    int findShortestPath(int s, int d);
  
    // print shortest path from a source vertex ‘s’ to
    // destination vertex ‘d’.
    int printShortestPath(int parent[], int s, int d);
};
  
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[2*V];
}
  
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;
            }
        }
    }
}
  
// Driver program to test methods of graph class
int main()
{
    // Create a graph given in the above diagram
    int V = 4;
    Graph g(V);
    g.addEdge(0, 1, 2);
    g.addEdge(0, 2, 2);
    g.addEdge(1, 2, 1);
    g.addEdge(1, 3, 1);
    g.addEdge(2, 0, 1);
    g.addEdge(2, 3, 2);
    g.addEdge(3, 3, 2);
  
    int src = 0, dest = 3;
    cout << "\nShortest Distance between " << src
         << " and " << dest << " is "
         << g.findShortestPath(src, dest);
  
    return 0;
}

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Python

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'''  Program to shortest path from a given source vertex s to
    a given destination vertex t.  Expected time complexity
    is O(V+E)'''
from collections import defaultdict
   
#This class represents a directed graph using adjacency list representation
class Graph:
   
    def __init__(self,vertices):
        self.V = vertices #No. of vertices
        self.V_org = vertices
        self.graph = defaultdict(list) # default dictionary to store graph
   
    # function to add an edge to graph
    def addEdge(self,u,v,w):
        if w == 1:
            self.graph[u].append(v)
        else:   
            '''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.'''
            self.graph[u].append(self.V)
            self.graph[self.V].append(v)
            self.V = self.V + 1
      
    # To print the shortest path stored in parent[]
    def printPath(self, parent, j):
        Path_len = 1
        if parent[j] == -1 and j < self.V_org : #Base Case : If j is source
            print j,
            return 0 # when parent[-1] then path length = 0 
        l = self.printPath(parent , parent[j])
  
        #incerement path length
        Path_len = l + Path_len
  
        # print node only if its less than original node length.
        # i.e do not print any new node that has been added later
        if j < self.V_org : 
            print j,
  
        return Path_len
  
    ''' This function mainly does BFS and prints the
        shortest path from src to dest. It is assumed
        that weight of every edge is 1'''
    def findShortestPath(self,src, dest):
  
        # Mark all the vertices as not visited
        # Initialize parent[] and visited[]
        visited =[False]*(self.V)
        parent =[-1]*(self.V)
   
        # Create a queue for BFS
        queue=[]
   
        # Mark the source node as visited and enqueue it
        queue.append(src)
        visited[src] = True
   
        while queue :
              
            # Dequeue a vertex from queue 
            s = queue.pop(0)
              
            # if s = dest then print the path and return
            if s == dest:
                return self.printPath(parent, s)
                  
   
            # 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 in self.graph[s]:
                if visited[i] == False:
                    queue.append(i)
                    visited[i] = True
                    parent[i] = s
   
   
# Create a graph given in the above diagram
g = Graph(4)
g.addEdge(0, 1, 2)
g.addEdge(0, 2, 2)
g.addEdge(1, 2, 1)
g.addEdge(1, 3, 1)
g.addEdge(2, 0, 1)
g.addEdge(2, 3, 2)
g.addEdge(3, 3, 2)
  
src = 0; dest = 3
print ("Shortest Path between %d and %d is  " %(src, dest)),
l = g.findShortestPath(src, dest)
print ("\nShortest Distance between %d and %d is %d " %(src, dest, l)),
  
#This code is contributed by Neelam Yadav

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Output :

Shortest Path between 0 and 3 is 0 1 3 
Shortest Distance between 0 and 3 is 3

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).

This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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