Find K vertices in the graph which are connected to at least one of remaining vertices
Given a connected graph with N vertices. The task is to select k(k must be less than or equals to n/2, not necessarily minimum) vertices from the graph such that all these selected vertices are connected to at least one of the non selected vertex. In case of multiple answers print any one of them.
Examples:
Input :
Output : 1
Vertex 1 is connected to all other non selected vertices. Here
{1, 2}, {2, 3}, {3, 4}, {1, 3}, {1, 4}, {2, 4} are also the valid answersInput :
Output : 1 3
Vertex 1, 3 are connected to all other non selected vertices. {2, 4} is also a valid answer.
Efficient Approach: An efficient way is to find vertices which are even level and odd level using simple dfs or bfs function. Then if the verices at odd level are less than the vertices at even level then print odd level vertices. Otherwise, print even level vertices.
Below is the implementation of the above approach:
C++
// C++ program to find K vertices in // the graph which are connected to at // least one of remaining vertices #include <bits/stdc++.h> using namespace std; #define N 200005 // To store graph int n, m, vis[N]; vector<int> gr[N]; vector<int> v[2]; // Function to add edge void add_edges(int x, int y) { gr[x].push_back(y); gr[y].push_back(x); } // Function to find level of each node void dfs(int x, int state) { // Push the vertex in respected level v[state].push_back(x); // Make vertex visited vis[x] = 1; // Traverse for all it's child nodes for (auto i : gr[x]) if (vis[i] == 0) dfs(i, state ^ 1); } // Function to print vertices void Print_vertices() { // If odd level vertices are less if (v[0].size() < v[1].size()) { for (auto i : v[0]) cout << i << " "; } // If even level vertices are less else { for (auto i : v[1]) cout << i << " "; } } // Driver code int main() { int n = 4, m = 3; // Add edges add_edges(1, 2); add_edges(2, 3); add_edges(3, 4); // Function call dfs(1, 0); Print_vertices(); return 0; } |
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Java
// Java program to find K vertices in // the graph which are connected to at // least one of remaining vertices import java.util.*; class GFG { static final int N = 200005; // To store graph static int n, m; static int[] vis = new int[N]; static Vector<Integer>[] gr = new Vector[N]; static Vector<Integer>[] v = new Vector[2]; // Function to add edge static void add_edges(int x, int y) { gr[x].add(y); gr[y].add(x); } // Function to find level of each node static void dfs(int x, int state) { // Push the vertex in respected level v[state].add(x); // Make vertex visited vis[x] = 1; // Traverse for all it's child nodes for (int i : gr[x]) { if (vis[i] == 0) { dfs(i, state ^ 1); } } } // Function to print vertices static void Print_vertices() { // If odd level vertices are less if (v[0].size() < v[1].size()) { for (int i : v[0]) { System.out.print(i + " "); } } // If even level vertices are less else { for (int i : v[1]) { System.out.print(i + " "); } } } // Driver code public static void main(String[] args) { n = 4; m = 3; for (int i = 0; i < N; i++) { gr[i] = new Vector<Integer>(); } for (int i = 0; i < 2; i++) { v[i] = new Vector<Integer>(); } // Add edges add_edges(1, 2); add_edges(2, 3); add_edges(3, 4); // Function call dfs(1, 0); Print_vertices(); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 program to find K vertices in # the graph which are connected to at # least one of remaining vertices N = 200005 # To store graph n, m, =0,0vis=[0 for i in range(N)] gr=[[] for i in range(N)] v=[[] for i in range(2)] # Function to add edge def add_edges(x, y): gr[x].append(y) gr[y].append(x) # Function to find level of each node def dfs(x, state): # Push the vertex in respected level v[state].append(x) # Make vertex visited vis[x] = 1 # Traverse for all it's child nodes for i in gr[x]: if (vis[i] == 0): dfs(i, state ^ 1) # Function to prvertices def Print_vertices(): # If odd level vertices are less if (len(v[0]) < len(v[1])): for i in v[0]: print(i,end=" ") # If even level vertices are less else: for i in v[1]: print(i,end=" ") # Driver code n = 4m = 3 # Add edges add_edges(1, 2) add_edges(2, 3) add_edges(3, 4) # Function call dfs(1, 0) Print_vertices() # This code is contributed by mohit kumar 29 |
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C#
// C# program to find K vertices in // the graph which are connected to at // least one of remaining vertices using System; using System.Collections.Generic; class GFG { static readonly int N = 200005; // To store graph static int n, m; static int[] vis = new int[N]; static List<int>[] gr = new List<int>[N]; static List<int>[] v = new List<int>[2]; // Function to add edge static void add_edges(int x, int y) { gr[x].Add(y); gr[y].Add(x); } // Function to find level of each node static void dfs(int x, int state) { // Push the vertex in respected level v[state].Add(x); // Make vertex visited vis[x] = 1; // Traverse for all it's child nodes foreach (int i in gr[x]) { if (vis[i] == 0) { dfs(i, state ^ 1); } } } // Function to print vertices static void Print_vertices() { // If odd level vertices are less if (v[0].Count < v[1].Count) { foreach (int i in v[0]) { Console.Write(i + " "); } } // If even level vertices are less else { foreach (int i in v[1]) { Console.Write(i + " "); } } } // Driver code public static void Main(String[] args) { n = 4; m = 3; for (int i = 0; i < N; i++) { gr[i] = new List<int>(); } for (int i = 0; i < 2; i++) { v[i] = new List<int>(); } // Add edges add_edges(1, 2); add_edges(2, 3); add_edges(3, 4); // Function call dfs(1, 0); Print_vertices(); } } // This code is contributed by Rajput-Ji |
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Output:
2 4
Time Complexity : O(V+E)
Where V is the number of vertices and E is the number of edges in the graph.
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