Minimum edge reversals to make a root

Given a directed tree with V vertices and V-1 edges, we need to choose such a root (from given nodes from where we can reach to every other node) with a minimum number of edge reversal.
Examples:



In above tree, if we choose node 3 as our 
root then we need to reverse minimum number
of 3 edges to reach every other node, 
changed tree is shown on the right side.



We can solve this problem using DFS. we start dfs at any random node of given tree and at each node we store its distance from starting node assuming all edges as undirected and we also store number of edges which need to be reversed in the path from starting node to current node, let’s denote such edges as back edges so back edges are those which point towards the node in a path. With this dfs, we also calculate total number of edge reversals in the tree. After this computation, at each node we can calculate ‘number of edge reversal to reach every other node’ as follows,
Let total number of reversals in tree when some node is chosen as starting node for dfs is R then if we want to reach every other node from node i we need to reverse all back edges from path node i to starting node and we also need to reverse all other back edges other than node i to starting node path. First part will be (distance of node i from starting node – back edges count at node i) because we want to reverse edges in path from node i to starting node it will be total edges (i.e. distance) minus back edges from starting node to node i (i.e. back edge count at node i). The second part will be (total edge reversal or total back edges of tree R – back edge count of node i). After calculating this value at each node we will choose minimum of them as our result.
In below code, in the given edge direction weight 0 is added and in reverse direction weight 1 is added which is used to count reversal edges in dfs method.

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find min edge reversal to
// make every node reachable from root
#include <bits/stdc++.h>
using namespace std;
  
// method to dfs in tree and populates disRev values
int dfs(vector< pair<int, int> > g[],
        pair<int, int> disRev[], bool visit[], int u)
{
    // visit current node
    visit[u] = true;
    int totalRev = 0;
  
    // looping over all neighbors
    for (int i = 0; i < g[u].size(); i++)
    {
        int v = g[u][i].first;
        if (!visit[v])
        {
            // distance of v will be one more than distance of u
            disRev[v].first = disRev[u].first + 1;
  
            // initialize back edge count same as
            // parent node's count
            disRev[v].second = disRev[u].second;
  
            // if there is a reverse edge from u to i,
            // then only update
            if (g[u][i].second)
            {
                disRev[v].second = disRev[u].second + 1;
                totalRev++;
            }
            totalRev += dfs(g, disRev, visit, v);
        }
    }
  
    // return total reversal in subtree rooted at u
    return totalRev;
}
  
// method prints root and minimum number of edge reversal
void printMinEdgeReverseForRootNode(int edges[][2], int e)
{
    // number of nodes are one more than number of edges
    int V = e + 1;
  
    // data structure to store directed tree
    vector< pair<int, int> > g[V];
  
    // disRev stores two values - distance and back
    // edge count from root node
    pair<int, int> disRev[V];
  
    bool visit[V];
  
    int u, v;
    for (int i = 0; i < e; i++)
    {
        u = edges[i][0];
        v = edges[i][1];
  
        // add 0 weight in direction of u to v
        g[u].push_back(make_pair(v, 0));
  
        // add 1 weight in reverse direction
        g[v].push_back(make_pair(u, 1));
    }
  
    //    initialize all variables
    for (int i = 0; i < V; i++)
    {
        visit[i] = false;
        disRev[i].first = disRev[i].second = 0;
    }
  
    int root = 0;
  
    // dfs populates disRev data structure and
    // store total reverse edge counts
    int totalRev = dfs(g, disRev, visit, root);
  
    // UnComment below lines to print each node's
    // distance and edge reversal count from root node
    /*
    for (int i = 0; i < V; i++)
    {
        cout << i << " : " << disRev[i].first
              << " " << disRev[i].second << endl;
    }
    */
  
    int res = INT_MAX;
  
    // loop over all nodes to choose minimum edge reversal
    for (int i = 0; i < V; i++)
    {
        // (reversal in path to i) + (reversal
        // in all other tree parts)
        int edgesToRev = (totalRev - disRev[i].second) +
                         (disRev[i].first - disRev[i].second);
  
        // choose minimum among all values
        if (edgesToRev < res)
        {
            res = edgesToRev;
            root = i;
        }
    }
  
    // print the designated root and total
    // edge reversal made
    cout << root << " " << res << endl;
}
  
// Driver code to test above methods
int main()
{
    int edges[][2] =
    {
        {0, 1},
        {2, 1},
        {3, 2},
        {3, 4},
        {5, 4},
        {5, 6},
        {7, 6}
    };
    int e = sizeof(edges) / sizeof(edges[0]);
  
    printMinEdgeReverseForRootNode(edges, e);
    return 0;
}

chevron_right


Output:

3 3

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up


Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.