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Count of node sequences of length K consisting of at least one black edge

  • Last Updated : 30 Jun, 2021

Given a tree consisting of N nodes numbered from [1, N] and is colored either black(denoted by 1) or green(denoted by 0), the task is to count the number of sequences of length K [a1, a2, ..aK] such that the path taken between consecutive nodes is the shortest one and the edges covered consist of at least one black edge. Since the answer can be large, print it to modulo of 109+7.

Input: N = 4, K = 4 
1-2 0 
2-3 0 
2-4 0 
Output:
Explanation: 
Since there is no black edge in the tree. There are no such sequences.

Input: N = 3, K = 3 
1-2 1 
2-3 1 
Output: 24 
Explanation: 
All the 33 sequences except for (1, 1, 1), (2, 2, 2) and (3, 3, 3) are included in the answer. 

Approach: 
The idea is to count the number of sequences of length K such that no black edge is covered. Let the count be temp. Then (NK) – temp is the required answer. temp can be easily calculated by removing the black edges and then calculating the size of different components of the resultant graph. 

Follow the steps below: 



  1. Initialize the value of ans as NK.
  2. Construct a graph G by adding only green edges.
  3. Perform a DFS traversal of the graph and keep subtracting (sizeK) from the ans where size is the number of nodes in different components of the graph G.

Below is the implementation of the above approach:

C++




// C++ implementation of the
// above approach
#include <bits/stdc++.h>
#define int long long int
using namespace std;
 
const int mod = 1e9 + 7;
const int N = 1e5 + 1;
 
// Vector to store the graph
vector<int> G[N];
 
// Iterative Function to calculate
// (a<sup>b</sup>) % mod in O(log b)
int power(int a, int b) {
 
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    a = a % mod;
    if (a == 0)
 
      // In case x is divisible by p;
      return 0;
 
    while (b > 0) {
 
        // If a is odd,
        // multiply x with result
        if (b & 1)
            res = (res * a) % mod;
 
        // b must be even now
        b = b >> 1; // b = b/2
        a = (a * a) % mod;
    }
    return res;
}
 
// DFS traversal of the nodes
void dfs(int i, int& size,
         bool* visited) {
     
    // Mark the current
    // node as visited
    visited[i] = true;
     
    // Increment  the size of the
    // current component
    size++;
 
    // Recur for all the vertices
    // adjacent to this node
    for (auto nbr : G[i]) {
        if (!visited[nbr]) {
          dfs(nbr, size, visited);
        }
    }
}
 
// Function to calculate the
// final answer
void totalSequences(int& ans,
                    int n, int k)
{
    // Mark all the vertices as
    // not visited initially
    bool visited[n + 1];
    memset(visited, false,
           sizeof(visited));
 
  // Subtract (size^k) from the answer
  // for different components
    for (int i = 1; i <= n; i++) {
        if (!visited[i]) {
            int size = 0;
            dfs(i, size, visited);
            ans -= power(size, k);
            ans += mod;
            ans = ans % mod;
        }
    }
}
 
// Function to add edges to the graph
void addEdge(int x, int y, int color)
{
    // If the colour is green,
    // include it in the graph
    if (color == 0) {
        G[x].push_back(y);
        G[y].push_back(x);
    }
}
int32_t main()
{
    // Number of node in the tree
    int n = 3;
 
    // Size of sequence
    int k = 3;
 
    // Initialize the result as n^k
    int ans = power(n, k);
    /* 2
     /   \
    1     3
        
   Let us create binary tree as shown
   in above example */
    addEdge(1, 2, 1);
    addEdge(2, 3, 1);
 
    totalSequences(ans, n, k);
    cout << ans << endl;
}

Java




// Java implementation of the
// above approach
import java.util.*;
 
class GFG{
 
static int mod = (int)(1e9 + 7);
static int N = (int)(1e5 + 1);
static  int size;
static int ans;
 
// Vector to store the graph
@SuppressWarnings("unchecked")
static Vector<Integer> []G = new Vector[N];
 
// Iterative Function to calculate
// (a<sup>b</sup>) % mod in O(log b)
static int power(int a, int b)
{
     
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    a = a % mod;
     
    if (a == 0)
     
        // In case x is divisible by p;
        return 0;
 
    while (b > 0)
    {
         
        // If a is odd,
        // multiply x with result
        if (b % 2 == 1)
            res = (res * a) % mod;
 
        // b must be even now
        b = b >> 1; // b = b/2
        a = (a * a) % mod;
    }
    return res;
}
 
// DFS traversal of the nodes
static void dfs(int i, boolean []visited)
{
     
    // Mark the current
    // node as visited
    visited[i] = true;
     
    // Increment  the size of the
    // current component
    size++;
 
    // Recur for all the vertices
    // adjacent to this node
    for(int nbr : G[i])
    {
        if (!visited[nbr])
        {
            dfs(nbr, visited);
        }
    }
}
 
// Function to calculate the
// final answer
static void totalSequences(int n, int k)
{
     
    // Mark all the vertices as
    // not visited initially
    boolean []visited = new boolean[n + 1];
 
    // Subtract (size^k) from the answer
    // for different components
    for(int i = 1; i <= n; i++)
    {
        if (!visited[i])
        {
            size = 0;
            dfs(i, visited);
            ans -= power(size, k);
            ans += mod;
            ans = ans % mod;
        }
    }
}
 
// Function to add edges to the graph
static void addEdge(int x, int y, int color)
{
     
    // If the colour is green,
    // include it in the graph
    if (color == 0)
    {
        G[x].add(y);
        G[y].add(x);
    }
}
 
// Driver code
public static void main(String[] args)
{
    for(int i = 0; i < G.length; i++)
        G[i] = new Vector<Integer>();
         
    // Number of node in the tree
    int n = 3;
 
    // Size of sequence
    int k = 3;
 
    // Initialize the result as n^k
    ans = power(n, k);
     
    /* 2
     /   \
    1     3
    Let us create binary tree as shown
    in above example */
    addEdge(1, 2, 1);
    addEdge(2, 3, 1);
 
    totalSequences(n, k);
    System.out.print(ans + "\n");
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program for the above approach
mod = 1e9 + 7
N = 1e5 + 1
 
# List to store the graph
G = [0] * int(N)
 
# Iterative Function to calculate
# (a<sup>b</sup>) % mod in O(log b)
def power(a, b):
 
    # Initialize result
    res = 1
 
    # Update x if it is more than
    # or equal to p
    a = a % mod
 
    if a == 0:
         
        # In case x is divisible by p;
        return 0
 
    while b > 0:
 
        # If a is odd,
        # multiply x with result
        if b & 1:
            res = (res * a) % mod
 
        # b must be even now
        b = b >> 1 # b = b/2
        a = (a * a) % mod
 
    return res
 
# DFS traversal of the nodes
def dfs(i, size, visited):
 
    # Mark the current
    # node as visited
    visited[i] = True
 
    # Increment the size of the
    # current component
    size[0] += 1
 
    # Recur for all the vertices
    # adjacent to this node
    for nbr in range(G[i]):
        if not visited[nbr]:
            dfs(nbr, size, visited)
 
# Function to calculate the
# final answer
def totalSequences(ans, n, k):
 
    # Mark all the vertices as
    # not visited initially
    visited = [False] * (n + 1)
 
    # Subtract (size^k) from the answer
    # for different components
    for i in range(1, n + 1):
        if not visited[i]:
            size = [0]
            dfs(i, size, visited)
             
            ans[0] -= power(size[0], k)
            ans[0] += mod
            ans[0] = ans[0] % mod
 
# Function to add edges to the graph
def addEdge(x, y, color):
 
    # If the colour is green,
    # include it in the graph
    if color == 0:
        G[x].append(y)
        G[y].append(x)
 
# Driver code
if __name__ == '__main__':
 
    # Number of node in the tree
    n = 3
 
    # Size of sequence
    k = 3
 
    # Initialize the result as n^k
    ans = [power(n, k)]
 
    """
      2
     / \
    1   3
         
    Let us create binary tree as shown
    in above example
    """
    addEdge(1, 2, 1)
    addEdge(2, 3, 1)
 
    totalSequences(ans, n, k)
    print(int(ans[0]))
 
# This code is contributed by Shivam Singh

C#




// C# implementation of the
// above approach
using System;
using System.Collections.Generic;
class GFG{
 
static int mod = (int)(1e9 + 7);
static int N = (int)(1e5 + 1);
static  int size;
static int ans;
 
// List to store the graph
static List<int> []G =
       new List<int>[N];
 
// Iterative Function to
// calculate (a<sup>b</sup>)
// % mod in O(log b)
static int power(int a,
                 int b)
{
  // Initialize result
  int res = 1;
 
  // Update x if it is
  // more than or equal
  // to p
  a = a % mod;
 
  if (a == 0)
 
    // In case x is
    // divisible by p;
    return 0;
 
  while (b > 0)
  {
    // If a is odd,
    // multiply x
    // with result
    if (b % 2 == 1)
      res = (res * a) % mod;
 
    // b must be even now
    // b = b/2
    b = b >> 1;
     
    a = (a * a) % mod;
  }
  return res;
}
 
// DFS traversal of the nodes
static void dfs(int i,
                bool []visited)
{
  // Mark the current
  // node as visited
  visited[i] = true;
 
  // Increment  the size of the
  // current component
  size++;
 
  // Recur for all the vertices
  // adjacent to this node
  foreach(int nbr in G[i])
  {
    if (!visited[nbr])
    {
      dfs(nbr, visited);
    }
  }
}
 
// Function to calculate the
// readonly answer
static void totalSequences(int n,
                           int k)
{
  // Mark all the vertices as
  // not visited initially
  bool []visited = new bool[n + 1];
 
  // Subtract (size^k) from the
  // answer for different components
  for(int i = 1; i <= n; i++)
  {
    if (!visited[i])
    {
      size = 0;
      dfs(i, visited);
      ans -= power(size, k);
      ans += mod;
      ans = ans % mod;
    }
  }
}
 
// Function to add edges
// to the graph
static void addEdge(int x,
                    int y,
                    int color)
{
  // If the colour is green,
  // include it in the graph
  if (color == 0)
  {
    G[x].Add(y);
    G[y].Add(x);
  }
}
 
// Driver code
public static void Main(String[] args)
{
  for(int i = 0; i < G.Length; i++)
    G[i] = new List<int>();
 
  // Number of node
  // in the tree
  int n = 3;
 
  // Size of sequence
  int k = 3;
 
  // Initialize the
  // result as n^k
  ans = power(n, k);
 
  /*   2
     /   \
    1     3
    Let us create binary tree
    as shown in above example */
  addEdge(1, 2, 1);
  addEdge(2, 3, 1);
 
  totalSequences(n, k);
  Console.Write(ans + "\n");
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
    // Javascript implementation of the above approach
     
    let mod = (1e9 + 7);
    let N = (1e5 + 1);
    let size;
    let ans;
 
    // List to store the graph
    let G = new Array(N);
 
    // Iterative Function to
    // calculate (a<sup>b</sup>)
    // % mod in O(log b)
    function power(a, b)
    {
      // Initialize result
      let res = 1;
 
      // Update x if it is
      // more than or equal
      // to p
      a = a % mod;
 
      if (a == 0)
 
        // In case x is
        // divisible by p;
        return 0;
 
      while (b > 0)
      {
        // If a is odd,
        // multiply x
        // with result
        if (b % 2 == 1)
          res = (res * a) % mod;
 
        // b must be even now
        // b = b/2
        b = b >> 1;
 
        a = (a * a) % mod;
      }
      return res;
    }
 
    // DFS traversal of the nodes
    function dfs(i, visited)
    {
      // Mark the current
      // node as visited
      visited[i] = true;
 
      // Increment  the size of the
      // current component
      size++;
 
      // Recur for all the vertices
      // adjacent to this node
      for(let nbr = 0; nbr < G[i].length; nbr++)
      {
        if (!visited[G[i][nbr]])
        {
          dfs(G[i][nbr], visited);
        }
      }
    }
 
    // Function to calculate the
    // readonly answer
    function totalSequences(n, k)
    {
      // Mark all the vertices as
      // not visited initially
      let visited = new Array(n + 1);
      visited.fill(false);
 
      // Subtract (size^k) from the
      // answer for different components
      for(let i = 1; i <= n; i++)
      {
        if (!visited[i])
        {
          size = 0;
          dfs(i, visited);
          ans -= power(size, k);
          ans += mod;
          ans = ans % mod;
        }
      }
    }
 
    // Function to add edges
    // to the graph
    function addEdge(x, y, color)
    {
      // If the colour is green,
      // include it in the graph
      if (color == 0)
      {
        G[x].push(y);
        G[y].push(x);
      }
    }
     
    for(let i = 0; i < G.length; i++)
        G[i] = [];
  
    // Number of node
    // in the tree
    let n = 3;
 
    // Size of sequence
    let k = 3;
 
    // Initialize the
    // result as n^k
    ans = power(n, k);
 
    /*   2
           /   \
          1     3
          Let us create binary tree
          as shown in above example */
    addEdge(1, 2, 1);
    addEdge(2, 3, 1);
 
    totalSequences(n, k);
    document.write(ans + "</br>");
 
</script>
Output: 
24

Time Complexity: O(N), Since DFS traversal requires O(Vertices + Edges) == O(N + (N-1)) == O(N) ) complexity. 
 




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