Minimum edges to be added in a directed graph so that any node can be reachable from a given node
Given a directed graph and a node X. The task is to find the minimum number of edges that must be added to the graph such that any node can be reachable from the given node.
Examples:
Input: X = 0
Output: 3
Input: X = 4
Output: 1
Approach: First, let’s mark all the vertices reachable from X as good, using a simple DFS. Then, for each bad vertex (vertices which are not reachable from X) v, count the number of bad vertices reachable from v (it also can be done by simple DFS). Let this number be cntv. Now, iterate over all bad vertices in non-increasing order of cntv. For the current bad vertex v, if it is still not marked as good, run a DFS from it, marking all the reachable vertices as good, and increase the answer by 1 (in fact, we are implicitly adding the edge (s, v)). It can be proved that this solution gives an optimal answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; const int N = 5010; int n, x; vector< int > g[N]; // To check if the vertex has been // visited or not bool vis[N]; // To store if vertex is reachable // from source or not bool good[N]; int cnt; void ADD_EDGE( int u, int v) { g[u].push_back(v); } // Function to find all good vertices void dfs1( int v) { good[v] = true ; for ( auto to : g[v]) if (!good[to]) dfs1(to); } // Function to find cnt of all unreachable vertices void dfs2( int v) { vis[v] = true ; ++cnt; for ( auto to : g[v]) if (!vis[to] && !good[to]) dfs2(to); } // Function to return the minimum edges required int Minimum_Edges() { // Find all vertices reachable from the source dfs1(x); // To store all vertices with their cnt value vector<pair< int , int > > val; for ( int i = 0; i < n; ++i) { // If vertex is bad i.e. not reachable if (!good[i]) { cnt = 0; memset (vis, false , sizeof (vis)); // Find cnt of this vertex dfs2(i); val.push_back(make_pair(cnt, i)); } } // Sort all unreachable vertices in // non-decreasing order of their cnt values sort(val.begin(), val.end()); reverse(val.begin(), val.end()); // Find the minimum number of edges // needed to be added int ans = 0; for ( auto it : val) { if (!good[it.second]) { ++ans; dfs1(it.second); } } return ans; } // Driver code int main() { // Number of nodes and source node n = 5, x = 4; // Add edges to the graph ADD_EDGE(0, 1); ADD_EDGE(1, 2); ADD_EDGE(2, 3); ADD_EDGE(3, 0); cout << Minimum_Edges(); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // pair static class pair { int first,second; pair( int a, int b) { first = a; second = b; } } static int N = 5010 ; static int n, x; static Vector<Vector<Integer>> g = new Vector<Vector<Integer>>(); // To check if the vertex has been // visited or not static boolean vis[] = new boolean [N]; // To store if vertex is reachable // from source or not static boolean good[] = new boolean [N]; static int cnt; static void ADD_EDGE( int u, int v) { g.get(u).add(v); } // Function to find all good vertices static void dfs1( int v) { good[v] = true ; for ( int to = 0 ; to < g.get(v).size(); to++) if (!good[g.get(v).get(to)]) dfs1(g.get(v).get(to)); } // Function to find cnt of all unreachable vertices static void dfs2( int v) { vis[v] = true ; ++cnt; for ( int to = 0 ; to < g.get(v).size(); to++) if (!vis[g.get(v).get(to)] && !good[g.get(v).get(to)]) dfs2(g.get(v).get(to)); } // Function to return the minimum edges required static int Minimum_Edges() { // Find all vertices reachable from the source dfs1(x); // To store all vertices with their cnt value Vector<pair> val = new Vector<pair>(); for ( int i = 0 ; i < n; ++i) { // If vertex is bad i.e. not reachable if (!good[i]) { cnt = 0 ; for ( int j = 0 ; j < vis.length; j++) vis[j] = false ; // Find cnt of this vertex dfs2(i); val.add( new pair(cnt, i)); } } // Sort all unreachable vertices in // non-decreasing order of their cnt values Collections.sort(val, new Comparator<pair>() { public int compare(pair p1, pair p2) { return p1.first - p2.first; } }); Collections.reverse(val); // Find the minimum number of edges // needed to be added int ans = 0 ; for ( int it = 0 ; it < val.size(); it++) { if (!good[val.get(it).second]) { ++ans; dfs1(val.get(it).second); } } return ans; } // Driver code public static void main(String args[]) { // Number of nodes and source node n = 5 ; x = 4 ; for ( int i = 0 ; i < N + 1 ; i++) g.add( new Vector<Integer>()); // Add edges to the graph ADD_EDGE( 0 , 1 ); ADD_EDGE( 1 , 2 ); ADD_EDGE( 2 , 3 ); ADD_EDGE( 3 , 0 ); System.out.println( Minimum_Edges()); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the approach N = 5010 g = [[] for i in range (N)] # To check if the vertex # has been visited or not vis = [ False for i in range (N)] # To store if vertex is reachable # from source or not good = [ False for i in range (N)] def ADD_EDGE(u, v): g[u].append(v) # Function to find all good vertices def dfs1(v): good[v] = True for to in g[v]: if not good[to]: dfs1(to) # Function to find cnt of # all unreachable vertices def dfs2(v): global cnt vis[v] = True cnt + = 1 for to in g[v]: if not vis[to] and not good[to]: dfs2(to) # Function to return # the minimum edges required def Minimum_Edges(): global vis, cnt # Find all vertices reachable # from the source dfs1(x) # To store all vertices # with their cnt value val = [] for i in range ( 0 , n): # If vertex is bad i.e. not reachable if not good[i]: cnt = 0 vis = [ False for i in range (N)] # Find cnt of this vertex dfs2(i) val.append((cnt, i)) # Sort all unreachable vertices # in non-decreasing order of # their cnt values val.sort(reverse = True ) # Find the minimum number of edges # needed to be added ans = 0 for it in val: if not good[it[ 1 ]]: ans + = 1 dfs1(it[ 1 ]) return ans # Driver code if __name__ = = "__main__" : # Number of nodes and source node n, x = 5 , 4 # Add edges to the graph ADD_EDGE( 0 , 1 ) ADD_EDGE( 1 , 2 ) ADD_EDGE( 2 , 3 ) ADD_EDGE( 3 , 0 ) print (Minimum_Edges()) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approach using System; using System.Collections; using System.Collections.Generic; class GFG { // pair class pair { public int first,second; public pair( int a, int b) { first = a; second = b; } } static int N = 5010; static int n, x; static ArrayList g = new ArrayList(); // To check if the vertex has been // visited or not static bool []vis = new bool [N]; // To store if vertex is reachable // from source or not static bool []good = new bool [N]; static int cnt; static void Add_EDGE( int u, int v) { ((ArrayList)g[u]).Add(v); } // Function to find all good vertices static void dfs1( int v) { good[v] = true ; for ( int to = 0; to < ((ArrayList)g[v]).Count; to++) if (!good[( int )((ArrayList)g[v])[to]]) dfs1(( int )((ArrayList)g[v])[to]); } // Function to find cnt of all unreachable vertices static void dfs2( int v) { vis[v] = true ; ++cnt; for ( int to = 0; to < ((ArrayList)g[v]).Count; to++) if (!vis[( int )((ArrayList)g[v])[to]] && !good[( int )((ArrayList)g[v])[to]]) dfs2(( int )((ArrayList)g[v])[to]); } class sortHelper : IComparer { int IComparer.Compare( object a, object b) { pair first = (pair)a; pair second = (pair)b; return first.first - second.first; } } // Function to return the minimum edges required static int Minimum_Edges() { // Find all vertices reachable from the source dfs1(x); // To store all vertices with their cnt value ArrayList val = new ArrayList(); for ( int i = 0; i < n; ++i) { // If vertex is bad i.e. not reachable if (!good[i]) { cnt = 0; for ( int j = 0; j < vis.Length; j++) vis[j] = false ; // Find cnt of this vertex dfs2(i); val.Add( new pair(cnt, i)); } } // Sort all unreachable vertices in // non-decreasing order of their cnt values val.Sort( new sortHelper()); // Find the minimum number of edges // needed to be Added int ans = 0; for ( int it = 0; it < val.Count; it++) { if (!good[((pair)val[it]).second]) { ++ans; dfs1(((pair)val[it]).second); } } return ans; } // Driver code public static void Main( string []args) { // Number of nodes and source node n = 5; x = 4; for ( int i = 0; i < N + 1; i++) g.Add( new ArrayList()); // Add edges to the graph Add_EDGE(0, 1); Add_EDGE(1, 2); Add_EDGE(2, 3); Add_EDGE(3, 0); Console.WriteLine(Minimum_Edges()); } } // This code is contributed by rutvik_56 |
Javascript
<script> // Javascript implementation of the approach class pair { constructor(a,b) { this .first=a; this .second=b; } } let N = 5010; let n, x; let g = []; // To check if the vertex has been // visited or not let vis= new Array(N); // To store if vertex is reachable // from source or not let good= new Array(N); for (let i=0;i<N;i++) { vis[i]= false ; good[i]= false ; } let cnt; function ADD_EDGE(u,v) { g[u].push(v); } // Function to find all good vertices function dfs1(v) { good[v] = true ; for (let to = 0; to < g[v].length; to++) if (!good[g[v][to]]) dfs1(g[v][to]); } // Function to find cnt of all unreachable vertices function dfs2(v) { vis[v] = true ; ++cnt; for (let to = 0; to < g[v].length; to++) if (!vis[g[v][to]] && !good[g[v][to]]) dfs2(g[v][to]); } // Function to return the minimum edges required function Minimum_Edges() { // Find all vertices reachable from the source dfs1(x); // To store all vertices with their cnt value let val = []; for (let i = 0; i < n; ++i) { // If vertex is bad i.e. not reachable if (!good[i]) { cnt = 0; for (let j = 0; j < vis.length; j++) vis[j] = false ; // Find cnt of this vertex dfs2(i); val.push( new pair(cnt, i)); } } // Sort all unreachable vertices in // non-decreasing order of their cnt values val.sort( function (p1,p2) { return p1.first - p2.first; } ); val.reverse(); // Find the minimum number of edges // needed to be added let ans = 0; for (let it = 0; it < val.length; it++) { if (!good[val[it].second]) { ++ans; dfs1(val[it].second); } } return ans; } // Driver code // Number of nodes and source node n = 5; x = 4; for (let i = 0; i < N + 1; i++) g.push([]); // Add edges to the graph ADD_EDGE(0, 1); ADD_EDGE(1, 2); ADD_EDGE(2, 3); ADD_EDGE(3, 0); document.write( Minimum_Edges()); // This code is contributed by avanitrachhadiya2155 </script> |
1
Time Complexity: O(V+E)
The time complexity of the above solution is O(V+E), where V is the number of vertices and E is the number of edges in the graph. We traverse over all the edges and vertices of the graph.
Space Complexity: O(V+E)
The space complexity of the above solution is also O(V+E), as we need to store all the edges and vertices of the graph.