Find two disjoint good sets of vertices in a given graph

Given an undirected unweighted graph with N vertices and M edges. The task is to find two disjoint good sets of vertices. A set X is called good if for every edge UV in the graph at least one of the endpoint belongs to X(i.e, U or V or both U and V belongs to X).
If it is not possible to make such sets then print -1.

Examples:

Input :

Output : {1 3 4 5} ,{2 6}
One disjoint good set contains vertices {1, 3, 4, 5} and other contains {2, 6}.



Input :

Output : -1

Approach:
One of the observation is that there is no edge UV that U and V are in the same set.The two good sets form a bipartition of the graph, so the graph has to be bipartite. And being bipartite is also sufficient. Read about bipartition here.

Below is the implementation of the above approach :

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to find two disjoint
// good sets of vertices in a given graph
#include <bits/stdc++.h>
using namespace std;
#define N 100005
  
// For the graph
vector<int> gr[N], dis[2];
bool vis[N];
int colour[N];
bool bip;
  
// Function to add edge
void Add_edge(int x, int y)
{
    gr[x].push_back(y);
    gr[y].push_back(x);
}
  
// Bipartie function
void dfs(int x, int col)
{
    vis[x] = true;
    colour[x] = col;
  
    // Check for child vertices
    for (auto i : gr[x]) {
  
        // If it is not visited
        if (!vis[i])
            dfs(i, col ^ 1);
  
        // If it is already visited
        else if (colour[i] == col)
            bip = false;
    }
}
  
// Function to find two disjoint
// good sets of vertices in a given graph
void goodsets(int n)
{
    // Initially assume that graph is bipartie
    bip = true;
  
    // For every unvisited vertex call dfs
    for (int i = 1; i <= n; i++)
        if (!vis[i])
            dfs(i, 0);
  
    // If graph is not bipartie
    if (!bip)
        cout << -1;
    else {
  
        // Differentiate two sets
        for (int i = 1; i <= n; i++)
            dis[colour[i]].push_back(i);
  
        // Print vertices belongs to both sets
  
        for (int i = 0; i < 2; i++) {
  
            for (int j = 0; j < dis[i].size(); j++)
                cout << dis[i][j] << " ";
            cout << endl;
        }
    }
}
  
// Driver code
int main()
{
    int n = 6, m = 4;
  
    // Add edges
    Add_edge(1, 2);
    Add_edge(2, 3);
    Add_edge(2, 4);
    Add_edge(5, 6);
  
    // Function call
    goodsets(n);
}

chevron_right






                        filter_none









Python 3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# Python 3 program to find two disjoint 


# good sets of vertices in a given graph 


N = 100005


  


# For the graph 


gr = [[] for i in range(N)] 


dis = [[] for i in range(2)] 


vis = [False for i in range(N)] 


colour = [0 for i in range(N)] 


bip = 0


  


# Function to add edge 


def Add_edge(x, y): 


    gr[x].append(y) 


    gr[y].append(x) 


  


# Bipartie function 


def dfs(x, col): 


    vis[x] = True


    colour[x] = col 


  


    # Check for child vertices 


    for i in gr[x]: 


          


        # If it is not visited 


        if (vis[i] == False): 


            dfs(i, col ^ 1) 


  


        # If it is already visited 


        elif (colour[i] == col): 


            bip = False


  


# Function to find two disjoint 


# good sets of vertices in a given graph 


def goodsets(n): 


      


    # Initially assume that  


    # graph is bipartie 


    bip = True


  


    # For every unvisited vertex call dfs 


    for i in range(1, n + 1, 1): 


        if (vis[i] == False): 


            dfs(i, 0) 


  


    # If graph is not bipartie 


    if (bip == 0): 


        print(-1) 


    else: 


          


        # Differentiate two sets 


        for i in range(1, n + 1, 1): 


            dis[colour[i]].append(i) 


  


        # Print vertices belongs to both sets 


        for i in range(2): 


            for j in range(len(dis[i])): 


                print(dis[i][j], end = " ")  


            print('\n', end = "") 


  


# Driver code 


if __name__ == '__main__': 


    n = 6


    m = 4


  


    # Add edges 


    Add_edge(1, 2) 


    Add_edge(2, 3) 


    Add_edge(2, 4) 


    Add_edge(5, 6) 


  


    # Function call 


    goodsets(n) 


  


# This code is contributed 


# by Surendra_Gangwar 



















                        chevron_right







                        filter_none











Output:


1 3 4 5 
2 6






          
          
          

          

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
pawan_asipu
pawanasipugmailcom
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 Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  SURENDRA_GANGWAR