Minimum edges to reverse to make path from a source to a destination

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

// Program to find minimum edge reversal to get 
// atleast one path from source to destination 
#include <bits/stdc++.h> 
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<int, int> > *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<int> shortestPath(int s); 
}; 
  
// Allocates memory for adjacency list 
Graph::Graph(int V) 
{ 
    this->V = V; 
    adj = new list< pair<int, int> >[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<int> Graph::shortestPath(int src) 
{ 
    // Create a set to store vertices that are being 
    // prerocessed 
    set< pair<int, int> > setds; 
  
    // Create a vector for distances and initialize all 
    // distances as infinite (INF) 
    vector<int> 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<int, int> 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<int, int> >::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<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 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 
} 

Output:

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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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