Convert Directed Graph into a Tree

Given an array arr[] of size N. There is an edge from i to arr[i]. The task is to convert this directed graph into tree by changing some of the edges. If for some i, arr[i] = i then i represents the root of the tree. In case of multiple answers print any of them.

Examples:

Input: arr[] = {6, 6, 0, 1, 4, 3, 3, 4, 0}
Output: {6, 6, 0, 1, 4, 3, 4, 4, 0}



Input: arr[] = {1, 2, 0};
Output: {0, 2, 0}.

Approach: Consider the 2nd example image above which shows an example of a functional graph. It consists of two cycles 1, 6, 3 and 4. Our goal is to make the graph consisting of exactly one cycle of exactly one vertex looped to itself.

Operation of change is equivalent to removing some outgoing edge and adding a new one, going to somewhat else vertex. Let’s firstly make our graph containing only one cycle. To do so, one can choose any of initially presented cycles and say that it will be the only one. Then one should consider every other cycle, remove any of its in-cycle edges and replace it with an edge going to any of the chosen cycle’s vertices. Thus the cycle will be broken and its vertices (along with tree ones) will be connected to the only chosen cycle. One will need to do exactly cycleCount – 1 such operations. Note that the removing of any non-cycle edge does not make sense, because it does not break any cycle.

The next thing is to make the cycle length be equal to 1. That might be already satisfied, if one will choose a cycle of minimal length and this length equals 1. Thus, if the initial graph contains any cycle of length 1, we are done with cycleCount – 1 operations. Otherwise, the cycle contains more than one vertex. It can be fixed with exactly one operation – one just needs to break any of in-cycle edges, say from u to arr[u], and add an edge from u to u. The graph will remain consisting of one cycle, but consisting of one self-looped vertex. In that case, we are done with cycleCount operations.

To do all the operations above, one can use DSU structure, or just a series of DFS. Note that there is no need in realisation of edge removing and creating, one just needs to analyze initial graph.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to convert directed graph into tree
#include <bits/stdc++.h>
using namespace std;
  
// Function to find root of the vertex
int find(int x, int a[], int vis[], int root[])
{
    if (vis[x])
        return root[x];
  
    vis[x] = 1;
    root[x] = x;
    root[x] = find(a[x], a, vis, root);
    return root[x];
}
  
// Function to convert directed graph into tree
void Graph_to_tree(int a[], int n)
{
    // Vis array to check if an index is visited or not
    // root[] array is to store parent of each vertex
    int vis[n] = { 0 }, root[n] = { 0 };
  
    // Find parent of each parent
    for (int i = 0; i < n; i++)
        find(a[i], a, vis, root);
  
    // par stores the root of the resulting tree
    int par = -1;
    for (int i = 0; i < n; i++)
        if (i == a[i])
            par = i;
  
    // If no self loop exists
    if (par == -1) {
        for (int i = 0; i < n; i++) {
  
            // Make vertex in a cycle as root of the tree
            if (i == find(a[i], a, vis, root)) {
                par = i;
                a[i] = i;
                break;
            }
        }
    }
  
    // Remove all cycles
    for (int i = 0; i < n; i++) {
        if (i == find(a[i], a, vis, root)) {
            a[i] = par;
        }
    }
  
    // Print the resulting array
    for (int i = 0; i < n; i++)
        cout << a[i] << " ";
}
  
// Driver code to test above functions
int main()
{
    int a[] = { 6, 6, 0, 1, 4, 3, 3, 4, 0 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    // Function call
    Graph_to_tree(a, n);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to convert
// directed graph into tree
import java.util.*;
      
class GFG
{
  
// Function to find root of the vertex
static int find(int x, int a[], 
                int vis[], int root[])
{
    if (vis[x] == 1)
        return root[x];
  
    vis[x] = 1;
    root[x] = x;
    root[x] = find(a[x], a, vis, root);
    return root[x];
}
  
// Function to convert directed graph into tree
static void Graph_to_tree(int a[], int n)
{
    // Vis array to check if an index is 
    // visited or not root[] array is to 
    // store parent of each vertex
    int []vis = new int[n];
    int []root = new int[n];
  
    // Find parent of each parent
    for (int i = 0; i < n; i++)
        find(a[i], a, vis, root);
  
    // par stores the root of the resulting tree
    int par = -1;
    for (int i = 0; i < n; i++)
        if (i == a[i])
            par = i;
  
    // If no self loop exists
    if (par == -1
    {
        for (int i = 0; i < n; i++) 
        {
  
            // Make vertex in a cycle as root of the tree
            if (i == find(a[i], a, vis, root))
            {
                par = i;
                a[i] = i;
                break;
            }
        }
    }
  
    // Remove all cycles
    for (int i = 0; i < n; i++)
    {
        if (i == find(a[i], a, vis, root))
        {
            a[i] = par;
        }
    }
  
    // Print the resulting array
    for (int i = 0; i < n; i++)
        System.out.print(a[i] + " ");
}
  
// Driver Code
static public void main ( String []arr)
{
    int a[] = { 6, 6, 0, 1, 4, 3, 3, 4, 0 };
  
    int n = a.length;
  
    // Function call
    Graph_to_tree(a, n);
}
}
  
// This code is contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# A Python3 program to convert 
# directed graph into tree
  
# Function to find root of the vertex
def find(x, a, vis, root):
    if vis[x]:
        return root[x]
  
    vis[x] = 1
    root[x] = x
    root[x] = find(a[x], a, vis, root)
    return root[x]
  
# Function to convert directed graph into tree
def Graph_To_Tree(a, n):
  
    # Vis array to check if an index is visited or not
    # root[] array is to store parent of each vertex
    vis = [0] * n
    root = [0] * n
  
    # Find parent of each parent
    for i in range(n):
        find(a[i], a, vis, root)
  
    # par stores the root of the resulting tree
    par = -1
    for i in range(n):
        if i == a[i]:
            par = i
  
    # If no self loop exists
    if par == -1:
        for i in range(n):
  
            # Make vertex in a cycle as root of the tree
            if i == find(a[i], a, vis, root):
                par = i
                a[i] = i
                break
  
    # Remove all cycles
    for i in range(n):
        if i == find(a[i], a, vis, root):
            a[i] = par
  
    # Print the resulting array
    for i in range(n):
        print(a[i], end = " ")
  
# Driver Code
if __name__ == "__main__":
    a = [6, 6, 0, 1, 4, 3, 3, 4, 0]
    n = len(a)
  
    # Function call
    Graph_To_Tree(a, n)
  
# This code is contributed by
# sanjeev2552

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to convert
// directed graph into tree
using System;
using System.Collections.Generic; 
  
class GFG
{
  
// Function to find root of the vertex
static int find(int x, int []a, 
                int []vis, int []root)
{
    if (vis[x] == 1)
        return root[x];
  
    vis[x] = 1;
    root[x] = x;
    root[x] = find(a[x], a, vis, root);
    return root[x];
}
  
// Function to convert directed graph into tree
static void Graph_to_tree(int []a, int n)
{
    // Vis array to check if an index is 
    // visited or not root[] array is to 
    // store parent of each vertex
    int []vis = new int[n];
    int []root = new int[n];
  
    // Find parent of each parent
    for (int i = 0; i < n; i++)
        find(a[i], a, vis, root);
  
    // par stores the root of the resulting tree
    int par = -1;
    for (int i = 0; i < n; i++)
        if (i == a[i])
            par = i;
  
    // If no self loop exists
    if (par == -1) 
    {
        for (int i = 0; i < n; i++) 
        {
  
            // Make vertex in a cycle as root of the tree
            if (i == find(a[i], a, vis, root))
            {
                par = i;
                a[i] = i;
                break;
            }
        }
    }
  
    // Remove all cycles
    for (int i = 0; i < n; i++)
    {
        if (i == find(a[i], a, vis, root))
        {
            a[i] = par;
        }
    }
  
    // Print the resulting array
    for (int i = 0; i < n; i++)
        Console.Write(a[i] + " ");
}
  
// Driver Code
static public void Main ( String []arr)
{
    int []a = { 6, 6, 0, 1, 4, 3, 3, 4, 0 };
  
    int n = a.Length;
  
    // Function call
    Graph_to_tree(a, n);
}
}
  
// This code is contributed by Princi Singh

chevron_right


Output:

6 6 0 1 4 3 4 4 0


My Personal Notes arrow_drop_up

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.





Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.