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.
Implementation:
// 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;
} |
// Java program to find min edge reversal to // make every node reachable from root import java.util.*;
class GFG
{ // pair class
static class pair
{
int first,second;
pair( int a , int b)
{
first = a;
second = b;
}
}
// method to dfs in tree and populates disRev values static int dfs(Vector<Vector< pair >> g,
pair disRev[], boolean visit[], int u)
{ // visit current node
visit[u] = true ;
int totalRev = 0 ;
// looping over all neighbors
for ( int i = 0 ; i < g.get(u).size(); i++)
{
int v = g.get(u).get(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.get(u).get(i).second!= 0 )
{
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 static void printMinEdgeReverseForRootNode( int edges[][], int e)
{ // number of nodes are one more than number of edges
int V = e + 1 ;
// data structure to store directed tree
Vector<Vector< pair >> g= new Vector<Vector< pair >>();
for ( int i = 0 ; i < V + 1 ; i++)
g.add( new Vector<pair>());
// disRev stores two values - distance and back
// edge count from root node
pair disRev[] = new pair[V];
for ( int i = 0 ; i < V; i++)
disRev[i] = new pair( 0 , 0 );
boolean visit[] = new boolean [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.get(u).add( new pair(v, 0 ));
// add 1 weight in reverse direction
g.get(v).add( new 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 = Integer.MAX_VALUE;
// 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
System.out.println(root + " " + res );
} // Driver code public static void main(String args[])
{ int edges[][] =
{
{ 0 , 1 },
{ 2 , 1 },
{ 3 , 2 },
{ 3 , 4 },
{ 5 , 4 },
{ 5 , 6 },
{ 7 , 6 }
};
int e = edges.length;
printMinEdgeReverseForRootNode(edges, e);
} } // This code is contributed by Arnab Kundu |
# Python3 program to find min edge reversal # to make every node reachable from root import sys
# Method to dfs in tree and populates # disRev values def dfs(g, disRev, visit, u):
# Visit current node
visit[u] = True
totalRev = 0
# Looping over all neighbors
for i in range ( len (g[u])):
v = g[u][i][ 0 ]
if ( not visit[v]):
# Distance of v will be one more
# than distance of u
disRev[v][ 0 ] = disRev[u][ 0 ] + 1
# Initialize back edge count same as
# parent node's count
disRev[v][ 1 ] = disRev[u][ 1 ]
# If there is a reverse edge from u to i,
# then only update
if (g[u][i][ 1 ]):
disRev[v][ 1 ] = disRev[u][ 1 ] + 1
totalRev + = 1
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 def printMinEdgeReverseForRootNode(edges, e):
# Number of nodes are one more than
# number of edges
V = e + 1
# Data structure to store directed tree
g = [[] for i in range (V)]
# disRev stores two values - distance
# and back edge count from root node
disRev = [[ 0 , 0 ] for i in range (V)]
visit = [ False for i in range (V)]
# u, v
for i in range (e):
u = edges[i][ 0 ]
v = edges[i][ 1 ]
# Add 0 weight in direction of u to v
g[u].append([v, 0 ])
# Add 1 weight in reverse direction
g[v].append([u, 1 ])
# Initialize all variables
for i in range (V):
visit[i] = False
disRev[i][ 0 ] = disRev[i][ 1 ] = 0
root = 0
# dfs populates disRev data structure and
# store total reverse edge counts
totalRev = dfs(g, disRev, visit, root)
# UnComment below lines to preach node's
# distance and edge reversal count from root node
# for (i = 0 i < V i++)
# {
# cout << i << " : " << disRev[i][0]
# << " " << disRev[i][1] << endl
# }
res = sys.maxsize
# Loop over all nodes to choose
# minimum edge reversal
for i in range (V):
# (reversal in path to i) + (reversal
# in all other tree parts)
edgesToRev = ((totalRev - disRev[i][ 1 ]) +
(disRev[i][ 0 ] - disRev[i][ 1 ]))
# Choose minimum among all values
if (edgesToRev < res):
res = edgesToRev
root = i
# Print the designated root and total
# edge reversal made
print (root, res)
# Driver code if __name__ = = '__main__' :
edges = [ [ 0 , 1 ], [ 2 , 1 ],
[ 3 , 2 ], [ 3 , 4 ],
[ 5 , 4 ], [ 5 , 6 ],
[ 7 , 6 ] ]
e = len (edges)
printMinEdgeReverseForRootNode(edges, e)
# This code is contributed by mohit kumar 29 |
// C# program to find min edge reversal to // make every node reachable from root using System;
using System.Collections.Generic;
class GFG
{ // pair class
public class pair
{
public int first,second;
public pair( int a, int b)
{
first = a;
second = b;
}
}
// method to dfs in tree and populates disRev values static int dfs(List<List< pair >> g,
pair []disRev, Boolean []visit, int u)
{ // visit current node
visit[u] = true ;
int totalRev = 0;
// looping over all neighbors
for ( int i = 0; i < g[u].Count; 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 != 0)
{
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 static void printMinEdgeReverseForRootNode( int [,]edges, int e)
{ // number of nodes are one more than number of edges
int V = e + 1;
// data structure to store directed tree
List<List< pair >> g = new List<List< pair >>();
for ( int i = 0; i < V + 1; i++)
g.Add( new List<pair>());
// disRev stores two values - distance and back
// edge count from root node
pair []disRev = new pair[V];
for ( int i = 0; i < V; i++)
disRev[i] = new pair(0, 0);
Boolean []visit = new Boolean[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].Add( new pair(v, 0));
// add 1 weight in reverse direction
g[v].Add( new 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 .MaxValue;
// 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
Console.WriteLine(root + " " + res);
} // Driver code public static void Main(String []args)
{ int [,]edges = {{0, 1}, {2, 1},
{3, 2}, {3, 4},
{5, 4}, {5, 6},
{7, 6}};
int e = edges.GetLength(0);
printMinEdgeReverseForRootNode(edges, e);
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program to find min edge // reversal to make every node reachable // from root class pair { constructor(a, b)
{
this .first = a;
this .second = b;
}
} // Method to dfs in tree and populates // disRev values function dfs(g, disRev, visit, u)
{ // Visit current node
visit[u] = true ;
let totalRev = 0;
// Looping over all neighbors
for (let i = 0; i < g[u].length; i++)
{
let 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 != 0)
{
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 function printMinEdgeReverseForRootNode(edges, e)
{ // Number of nodes are one more
// than number of edges
let V = e + 1;
// Data structure to store directed tree
let g = [];
for (let i = 0; i < V + 1; i++)
g.push([]);
// disRev stores two values - distance and
// back edge count from root node
let disRev = new Array(V);
for (let i = 0; i < V; i++)
disRev[i] = new pair(0, 0);
let visit = new Array(V);
let u, v;
for (let 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( new pair(v, 0));
// Add 1 weight in reverse direction
g[v].push( new pair(u, 1));
}
// Initialize all variables
for (let i = 0; i < V; i++)
{
visit[i] = false ;
disRev[i].first = disRev[i].second = 0;
}
let root = 0;
// dfs populates disRev data structure and
// store total reverse edge counts
let 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;
}
*/
let res = Number.MAX_VALUE;
// Loop over all nodes to choose
// minimum edge reversal
for (let i = 0; i < V; i++)
{
// (reversal in path to i) + (reversal
// in all other tree parts)
let 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
document.write(root + " " + res );
} // Driver code let edges = [ [ 0, 1 ], [ 2, 1 ], [ 3, 2 ], [ 3, 4 ],
[ 5, 4 ], [ 5, 6 ],
[ 7, 6 ] ];
let e = edges.length; printMinEdgeReverseForRootNode(edges, e); // This code is contributed by rag2127 </script> |
3 3
Space complexity of the given C++ program is O(V + E), where V is the number of nodes and E is the number of edges in the graph.
Time complexity of the program is O(V * E), where V is the number of nodes and E is the number of edges in the graph. This is because the dfs function is called for every node in the graph, and for each node, we loop over all its neighbors (which can be at most E). Therefore, the total number of iterations is V * E.