Number of ways to select a node from each connected component

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Given a graph with N nodes and M edges. The task is to find the number of ways to select a node from each connected component of the given graph.
Examples:

Input:

Output: 3
(1, 4), (2, 4), (3, 4) are possible ways.
Input:

Output: 6
(1, 4, 5), (2, 4, 5), (3, 4, 5), (1, 4, 6), (2, 4, 6), (3, 4, 6) are possible ways.

Approach: A product of the number of nodes in each connected component is the required answer. Run a simple dfs to find the number of nodes in each connected component.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100005
 
int n, m, temp;
vector<int> gr[N];
int vis[N];
 
// Function to add edges in the graph
void Add_edges(int x, int y)
{
    gr[x].push_back(y);
    gr[y].push_back(x);
}
 
// Function for DFS
void dfs(int ch)
{
    // Mark node as visited
    vis[ch] = 1;
 
    // Count number of nodes in a component
    temp++;
    for (auto i : gr[ch])
        if (!vis[i])
            dfs(i);
}
 
// Function to return the required number of ways
int NumberOfWays()
{
 
    // To store the required answer
    int ans = 1;
 
    memset(vis, 0, sizeof vis);
    for (int i = 1; i <= n; i++) {
 
        // If current node hasn't been visited yet
        if (!vis[i]) {
            temp = 0;
            dfs(i);
 
            // Multiply it with the answer
            ans *= temp;
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    n = 4, m = 2;
 
    // Add edges
    Add_edges(1, 2);
    Add_edges(1, 3);
 
    cout << NumberOfWays();
 
    return 0;
}

Java

// Java implementation of the approach 
import java.util.*;
 
class GFG
{ 
     
static final int N = 100005;
 
static int n, m, temp; 
static Vector<Integer> []gr = new Vector[N]; 
static int []vis = new int[N]; 
 
// Function to add edges in the graph 
static void Add_edges(int x, int y) 
{ 
    gr[x].add(y); 
    gr[y].add(x); 
} 
 
// Function for DFS 
static void dfs(int ch) 
{ 
    // Mark node as visited 
    vis[ch] = 1; 
 
    // Count number of nodes in a component 
    temp++; 
    for (int i : gr[ch]) 
        if (vis[i] == 0) 
            dfs(i); 
} 
 
// Function to return the required number of ways 
static int NumberOfWays() 
{ 
 
    // To store the required answer 
    int ans = 1; 
    Arrays.fill(vis, 0);
 
    for (int i = 1; i <= n; i++)
    { 
 
        // If current node hasn't been visited yet 
        if (vis[i] == 0) 
        { 
            temp = 0; 
            dfs(i); 
 
            // Multiply it with the answer 
            ans *= temp; 
        } 
    } 
 
    return ans; 
} 
 
// Driver code 
public static void main(String[] args) 
{ 
    n = 4;
    m = 2; 
    for (int i = 0; i < N; i++)
        gr[i] = new Vector<Integer>();
     
    // Add edges 
    Add_edges(1, 2); 
    Add_edges(1, 3); 
 
    System.out.print(NumberOfWays()); 
}
} 
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
 
# Function to add edges in the graph
def Add_edges(x, y):
 
    gr[x].append(y)
    gr[y].append(x)
 
# Function for DFS
def dfs(ch):
 
    # Mark node as visited
    vis[ch] = 1
    global temp
     
    # Count number of nodes 
    # in a component
    temp += 1
    for i in gr[ch]:
        if not vis[i]:
            dfs(i)
 
# Function to return the required 
# number of ways
def NumberOfWays():
 
    # To store the required answer
    ans = 1
    global temp
     
    for i in range(1, n + 1): 
 
        # If current node hasn't been 
        # visited yet
        if not vis[i]:
            temp = 0
            dfs(i)
 
            # Multiply it with the answer
            ans *= temp
         
    return ans
 
# Driver code
if __name__ == "__main__":
 
    n, m, temp = 4, 2, 0
    N = 100005
 
    gr = [[] for i in range(N)]
    vis = [None] * N
 
    # Add edges
    Add_edges(1, 2)
    Add_edges(1, 3)
 
    print(NumberOfWays())
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach 
using System;
using System.Collections.Generic;
 
class GFG
{ 
     
static readonly int N = 100005;
 
static int n, m, temp; 
static List<int> []gr = new List<int>[N]; 
static int []vis = new int[N]; 
 
// Function to add edges in the graph 
static void Add_edges(int x, int y) 
{ 
    gr[x].Add(y); 
    gr[y].Add(x); 
} 
 
// Function for DFS 
static void dfs(int ch) 
{ 
    // Mark node as visited 
    vis[ch] = 1; 
 
    // Count number of nodes in a component 
    temp++; 
    foreach (int i in gr[ch]) 
        if (vis[i] == 0) 
            dfs(i); 
} 
 
// Function to return the required number of ways 
static int NumberOfWays() 
{ 
 
    // To store the required answer 
    int ans = 1; 
    for(int i= 0; i < N; i++)
        vis[i] = 0;
 
    for (int i = 1; i <= n; i++)
    { 
 
        // If current node hasn't been visited yet 
        if (vis[i] == 0) 
        { 
            temp = 0; 
            dfs(i); 
 
            // Multiply it with the answer 
            ans *= temp; 
        } 
    } 
 
    return ans; 
} 
 
// Driver code 
public static void Main(String[] args) 
{ 
    n = 4;
    m = 2; 
    for (int i = 0; i < N; i++)
        gr[i] = new List<int>();
     
    // Add edges 
    Add_edges(1, 2); 
    Add_edges(1, 3); 
 
    Console.Write(NumberOfWays()); 
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of the approach 
 
let N = 100005;
 
let n, m, temp; 
 
let gr = new Array(N);
let vis = new Array(N);
 
// Function to add edges in the graph 
function Add_edges(x,y)
{
    gr[x].push(y); 
    gr[y].push(x);
}
 
// Function for DFS 
function dfs(ch)
{
    // Mark node as visited 
    vis[ch] = 1; 
   
    // Count number of nodes in a component 
    temp++; 
    for (let i of gr[ch]) 
        if (vis[i] == 0) 
            dfs(i); 
}
 
// Function to return the required number of ways 
function NumberOfWays() 
{
    // To store the required answer 
    let ans = 1; 
    for(let i=0;i<vis.length;i++)
        vis[i]=0;
   
    for (let i = 1; i <= n; i++)
    { 
   
        // If current node hasn't been visited yet 
        if (vis[i] == 0) 
        { 
            temp = 0; 
            dfs(i); 
   
            // Multiply it with the answer 
            ans *= temp; 
        } 
    } 
   
    return ans; 
}
 
// Driver code 
n = 4;
m = 2; 
for (let i = 0; i < N; i++)
    gr[i] = [];
 
// Add edges 
Add_edges(1, 2); 
Add_edges(1, 3); 
 
document.write(NumberOfWays()); 
 
// This code is contributed by unknown2108
</script>

Output:

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