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Minimum edges to reverse to make path from a source to a destination
  • Difficulty Level : Hard
  • Last Updated : 15 Mar, 2021
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Given a directed graph and a source node and destination node, we need to find how many edges we need to reverse in order to make at least 1 path from source node to destination node.

Examples:  

In above graph there were two paths from node 0 to node 6,
0 -> 1 -> 2 -> 3 -> 6
0 -> 1 -> 5 -> 4 -> 6
But for first path only two edges need to be reversed, so answer will be 2 only.

This problem can be solved assuming a different version of the given graph. In this version we make a reverse edge corresponding to every edge and we assign that a weight 1 and assign a weight 0 to original edge. After this modification above graph looks something like below, 
 



Now we can see that we have modified the graph in such a way that, if we move towards original edge, no cost is incurred, but if we move toward reverse edge 1 cost is added. So if we apply Dijkstra’s shortest path on this modified graph from given source, then that will give us minimum cost to reach from source to destination i.e. minimum edge reversal from source to destination. 

Below is the code based on above concept. 

C++




// C++ Program to find minimum edge reversal to get
// atleast one path from source to destination
#include
using namespace std;
# define INF 0x3f3f3f3f
 
// This class represents a directed graph using
// adjacency list representation
class Graph
{
    int V;    // No. of vertices
 
    // In a weighted graph, we need to store vertex
    // and weight pair for every edge
    list< pair > *adj;
 
public:
    Graph(int V);  // Constructor
 
    // function to add an edge to graph
    void addEdge(int u, int v, int w);
 
    // returns shortest path from s
    vector shortestPath(int s);
};
 
// Allocates memory for adjacency list
Graph::Graph(int V)
{
    this->V = V;
    adj = new list< pair >[V];
}
 
//  method adds a directed edge from u to v with weight w
void Graph::addEdge(int u, int v, int w)
{
    adj[u].push_back(make_pair(v, w));
}
 
// Prints shortest paths from src to all other vertices
vector Graph::shortestPath(int src)
{
    // Create a set to store vertices that are being
    // prerocessed
    set< pair > setds;
 
    // Create a vector for distances and initialize all
    // distances as infinite (INF)
    vector dist(V, INF);
 
    // Insert source itself in Set and initialize its
    // distance as 0.
    setds.insert(make_pair(0, src));
    dist[src] = 0;
 
    /* Looping till all shortest distance are finalized
       then setds will become empty */
    while (!setds.empty())
    {
        // The first vertex in Set is the minimum distance
        // vertex, extract it from set.
        pair tmp = *(setds.begin());
        setds.erase(setds.begin());
 
        // vertex label is stored in second of pair (it
        // has to be done this way to keep the vertices
        // sorted distance (distance must be first item
        // in pair)
        int u = tmp.second;
 
        // 'i' is used to get all adjacent vertices of a vertex
        list< pair >::iterator i;
        for (i = adj[u].begin(); i != adj[u].end(); ++i)
        {
            // Get vertex label and weight of current adjacent
            // of u.
            int v = (*i).first;
            int weight = (*i).second;
 
            //  If there is shorter path to v through u.
            if (dist[v] > dist[u] + weight)
            {
                /*  If distance of v is not INF then it must be in
                    our set, so removing it and inserting again
                    with updated less distance.
                    Note : We extract only those vertices from Set
                    for which distance is finalized. So for them,
                    we would never reach here.  */
                if (dist[v] != INF)
                    setds.erase(setds.find(make_pair(dist[v], v)));
 
                // Updating distance of v
                dist[v] = dist[u] + weight;
                setds.insert(make_pair(dist[v], v));
            }
        }
    }
    return dist;
}
 
/* method adds reverse edge of each original edge
   in the graph. It gives reverse edge a weight = 1
   and all original edges a weight of 0. Now, the
   length of the shortest path will give us the answer.
   If shortest path is p: it means we used p reverse
   edges in the shortest path. */
Graph modelGraphWithEdgeWeight(int edge[][2], int E, int V)
{
    Graph g(V);
    for (int i = 0; i < E; i++)
    {
        //  original edge : weight 0
        g.addEdge(edge[i][0], edge[i][1], 0);
 
        //  reverse edge : weight 1
        g.addEdge(edge[i][1], edge[i][0], 1);
    }
    return g;
}
 
// Method returns minimum number of edges to be
// reversed to reach from src to dest
int getMinEdgeReversal(int edge[][2], int E, int V,
                       int src, int dest)
{
    //  get modified graph with edge weight
    Graph g = modelGraphWithEdgeWeight(edge, E, V);
 
    //  get shortes path vector
    vector dist = g.shortestPath(src);
 
    // If distance of destination is still INF,
    // not possible
    if (dist[dest] == INF)
        return -1;
    else
        return dist[dest];
}
 
//  Driver code to test above method
int main()
{
    int V = 7;
    int edge[][2] = {{0, 1}, {2, 1}, {2, 3}, {5, 1},
                     {4, 5}, {6, 4}, {6, 3}};
    int E = sizeof(edge) / sizeof(edge[0]);
 
    int minEdgeToReverse =
                  getMinEdgeReversal(edge, E, V, 0, 6);
    if (minEdgeToReverse != -1)
        cout << minEdgeToReverse << endl;
    else
        cout << "Not possible" << endl;
    return 0;
}

Java




// Java program to find minimum edge reversal to get
// atleast one path from source to destination
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
 
class Pair
{
    int first, second;
 
    public Pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// This class represents a directed graph using
// adjacency list representation
class Graph{
     
final int INF = (int)0x3f3f3f3f;
 
// No. of vertices
int V;
 
// In a weighted graph, we need to store vertex
// and weight pair for every edge
List<Pair>[] adj;
 
// Allocates memory for adjacency list
@SuppressWarnings("unchecked")
public Graph(int V)
{
    this.V = V;
    adj = new ArrayList[V];
 
    for(int i = 0; i < V; i++)
    {
        adj[i] = new ArrayList();
    }
}
 
// Fucntion adds a directed edge from
// u to v with weight w
void addEdge(int u, int v, int w)
{
    adj[u].add(new Pair(v, w));
}
 
// Prints shortest paths from
// src to all other vertices
int[] shortestPath(int src)
{
     
    // Create a set to store vertices
    // that are being prerocessed
    Set<Pair> setds = new HashSet<Pair>();
 
    // Create a vector for distances and
    // initialize all distances as infinite(INF)
    int[] dist = new int[V];
    Arrays.fill(dist, INF);
 
    // Insert source itself in Set and initialize
    // its distance as 0.
    setds.add(new Pair(0, src));
    dist[src] = 0;
 
    // Looping till all shortest distance are
    // finalized then setds will become empty
    while (!setds.isEmpty())
    {
         
        // The first vertex in Set is the minimum
        // distance vertex, extract it from set.
        Iterator<Pair> itr = setds.iterator();
        Pair tmp = itr.next();
        itr.remove();
 
        // Vertex label is stored in second of pair (it
        // has to be done this way to keep the vertices
        // sorted distance (distance must be first item
        // in pair)
        int u = tmp.second;
 
        // 'i' is used to get all adjacent
        // vertices of a vertex
        for(Pair p : adj[u])
        {
             
            // Get vertex label and weight of
            // current adjacent of u.
            int v = p.first;
            int weight = p.second;
 
            // If there is shorter path to v through u.
            if (dist[v] > dist[u] + weight)
            {
                 
                // If distance of v is not INF then it
                // must be in our set, so removing it
                // and inserting again with updated
                // less distance. Note : We extract
                // only those vertices from Set for
                // which distance is finalized. So
                // for them, we would never reach here.
                if (dist[v] != INF)
                {
                    setds.remove(new Pair(dist[v], v));
                }
                 
                // setds.erase(setds.find(new Pair(dist[v], v)));
 
                // Updating distance of v
                dist[v] = dist[u] + weight;
                setds.add(new Pair(dist[v], v));
            }
        }
    }
    return dist;
}
}
 
class GFG{
static final int INF = (int)0x3f3f3f3f;
 
// Function adds reverse edge of each original
// edge in the graph. It gives reverse edge
// a weight = 1 and all original edges a
// weight of 0. Now, the length of the
// shortest path will give us the answer.
// If shortest path is p: it means we
// used p reverse edges in the shortest path.
static Graph modelGraphWithEdgeWeight(int edge[][],
                                      int E, int V)
{
    Graph g = new Graph(V);
    for(int i = 0; i < E; i++)
    {
         
        // Original edge : weight 0
        g.addEdge(edge[i][0], edge[i][1], 0);
 
        // Reverse edge : weight 1
        g.addEdge(edge[i][1], edge[i][0], 1);
    }
    return g;
}
 
// Function returns minimum number of edges to be
// reversed to reach from src to dest
static int getMinEdgeReversal(int edge[][], int E,
                              int V, int src, int dest)
{
     
    // Get modified graph with edge weight
    Graph g = modelGraphWithEdgeWeight(edge, E, V);
 
    // Get shortes path vector
    int[] dist = g.shortestPath(src);
 
    // If distance of destination is still INF,
    // not possible
    if (dist[dest] == INF)
        return -1;
    else
        return dist[dest];
}
 
// Driver code
public static void main(String[] args)
{
    int V = 7;
    int edge[][] = { { 0, 1 }, { 2, 1 },
                     { 2, 3 }, { 5, 1 },
                     { 4, 5 }, { 6, 4 },
                     { 6, 3 } };
    int E = edge.length;
 
    int minEdgeToReverse = getMinEdgeReversal(
        edge, E, V, 0, 6);
     
    if (minEdgeToReverse != -1)
        System.out.println(minEdgeToReverse);
    else
        System.out.println("Not possible");
}
}
 
// This code is contributed by sanjeev2552

Python3




# Python3 Program to find minimum edge reversal to get
# atleast one path from source to destination
 
# method adds a directed edge from u to v with weight w
def addEdge(u, v, w):
    global adj
    adj[u].append((v, w))
 
# Prints shortest paths from src to all other vertices
def shortestPath(src):
   
    # Create a set to store vertices that are being
    # prerocessed
    setds = {}
 
    # Create a vector for distances and initialize all
    # distances as infinite (INF)
    dist = [10**18 for i in range(V)]
 
    # Insert source itself in Set and initialize its
    global adj
    setds[(0, src)] = 1
    dist[src] = 0
 
    # /* Looping till all shortest distance are finalized
 
    # then setds will become empty */
    while (len(setds) > 0):
       
        # The first vertex in Set is the minimum distance
        # vertex, extract it from set.
        tmp = list(setds.keys())[0]
        del setds[tmp]
 
        # vertex label is stored in second of pair (it
        # has to be done this way to keep the vertices
        # sorted distance (distance must be first item
        # in pair)
        u = tmp[1]
 
        # 'i' is used to get all adjacent vertices of a vertex
        # list< pair<int, int> >::iterator i;
        for i in adj[u]:
           
            # Get vertex label and weight of current adjacent
            # of u.
            v = i[0];
            weight = i[1]
 
            # If there is shorter path to v through u.
            if (dist[v] > dist[u] + weight):
               
                # /* If distance of v is not INF then it must be in
                #     our set, so removing it and inserting again
                #     with updated less distance.
                #     Note : We extract only those vertices from Set
                #     for which distance is finalized. So for them,
                #     we would never reach here. */
                if (dist[v] != 10**18):
                    del setds[(dist[v], v)]
 
                # Updating distance of v
                dist[v] = dist[u] + weight
                setds[(dist[v], v)] = 1
 
    return dist
 
# /* method adds reverse edge of each original edge
# in the graph. It gives reverse edge a weight = 1
# and all original edges a weight of 0. Now, the
# length of the shortest path will give us the answer.
# If shortest path is p: it means we used p reverse
# edges in the shortest path. */
def modelGraphWithEdgeWeight(edge, E, V):
    global adj
    for i in range(E):
       
        # original edge : weight 0
        addEdge(edge[i][0], edge[i][1], 0)
 
        # reverse edge : weight 1
        addEdge(edge[i][1], edge[i][0], 1)
 
# Method returns minimum number of edges to be
# reversed to reach from src to dest
def getMinEdgeReversal(edge, E, V,src, dest):
   
    # get modified graph with edge weight
    modelGraphWithEdgeWeight(edge, E, V)
 
    # get shortes path vector
    dist = shortestPath(src)
 
    # If distance of destination is still INF,
    # not possible
    if (dist[dest] == 10**18):
        return -1
    else:
        return dist[dest]
 
# Driver code
if __name__ == '__main__':
    V = 7
    edge = [[0, 1], [2, 1], [2, 3], [5, 1],[4, 5], [6, 4], [6, 3]]
    E, adj = len(edge), [[] for i in range(V + 1)]
    minEdgeToReverse = getMinEdgeReversal(edge, E, V, 0, 6)
    if (minEdgeToReverse != -1):
        print(minEdgeToReverse)
    else:
        print("Not possible")
 
        # This code is contributed by mohit kumar 29

Output: 

2

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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