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: 0
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:
- Initialize the value of ans as NK.
- Construct a graph G by adding only green edges.
- 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++ 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 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 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# 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 |
<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> |
24
Time Complexity: O(N), Since DFS traversal requires O(Vertices + Edges) == O(N + (N-1)) == O(N) ) complexity.