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++

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// 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);
}

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Python 3

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# 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

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Output:

1 3 4 5 
2 6


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Improved By : SURENDRA_GANGWAR