Número de caminos de peso W en un árbol K-ario

Dado un árbol K-ario, donde cada Node tiene K hijos y cada borde tiene algo de peso. Todos los bordes, es decir, K, que van desde un Node en particular a todos sus hijos tienen pesos en orden ascendente 1, 2, 3, …, K. Encuentre el número de caminos que tienen un peso total como W (suma de todos los pesos de los bordes en el camino ) a partir de la raíz y que contiene al menos un borde de peso al menos M. 
Ejemplos: 
 

Input : W = 3, K = 3, M = 2
Output : 3
Explanation : One path can be (1 + 2), second can be (2 + 1) and third is 3.

Input : W = 4, K = 3, M = 2
Output : 6

Enfoque: Este problema se puede resolver utilizando el enfoque de programación dinámica. La idea es mantener dos estados, uno para que se requiera el peso actual y otro para una variable booleana que denota que la ruta actual ha incluido un borde de peso al menos M o no. Iterar sobre todos los pesos de borde posibles, es decir, K y resolver recursivamente para el peso W – i para 1 ≤ i ≤ K . Si el peso del borde actual es mayor o igual a M, configure la variable booleana como 1 para la siguiente llamada.
A continuación se muestra la implementación del enfoque anterior. 
 

// C++ program to count the number of
// paths with weight W in a K-ary tree
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number of ways
// having weight as wt in K-ary tree
int solve(int dp[][2], int wt, int K, int M,
          int used)
{
    // Return 0 if weight becomes less
    // than zero
    if (wt < 0)
        return 0;
 
    if (wt == 0) {
 
        // Return one only if the
        // current path has included
        // edge weight of atleast M
        if (used)
            return 1;
        return 0;
    }
 
    if (dp[wt][used] != -1)
        return dp[wt][used];
 
    int ans = 0;
    for (int i = 1; i <= K; i++) {
 
        // If the current edge weight
        // is greater than or equal to
        // M, set used as true
        if (i >= M)
            ans += solve(dp, wt - i,
                         K, M, used | 1);
        else
            ans += solve(dp, wt - i,
                         K, M, used);
    }
    return dp[wt][used] = ans;
}
 
// Driver Code to test above function
int main()
{
    int W = 3, K = 3, M = 2;
    int dp[W + 1][2];
    memset(dp, -1, sizeof(dp));
    cout << solve(dp, W, K, M, 0) << endl;
    return 0;
}

// Java program to count the number of
// paths with weight W in a K-ary tree
 
class GFG
{
    // Function to return the number of ways
    // having weight as wt in K-ary tree
 
    public static int solve(int[][] dp, int wt,
                            int K, int M, int used)
    {
        // Return 0 if weight becomes less
        // than zero
        if (wt < 0)
        {
            return 0;
        }
 
        if (wt == 0)
        {
 
            // Return one only if the
            // current path has included
            // edge weight of atleast M
            if (used == 1)
            {
                return 1;
            }
            return 0;
        }
 
        if (dp[wt][used] != -1)
        {
            return dp[wt][used];
        }
 
        int ans = 0;
        for (int i = 1; i <= K; i++)
        {
 
            // If the current edge weight
            // is greater than or equal to
            // M, set used as true
            if (i >= M)
            {
                ans += solve(dp, wt - i,
                        K, M, used | 1);
            }
            else
            {
                ans += solve(dp, wt - i,
                        K, M, used);
            }
        }
        return dp[wt][used] = ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int W = 3, K = 3, M = 2;
        int[][] dp = new int[W + 1][2];
        for (int i = 0; i < W + 1; i++)
        {
            for (int j = 0; j < 2; j++)
            {
                dp[i][j] = -1;
            }
        }
        System.out.print(solve(dp, W, K, M, 0) + "\n");
    }
}
 
// This code has been contributed by 29AjayKumar

# Python 3 program to count the number of
# paths with weight W in a K-ary tree
import numpy as np
 
# Function to return the number of ways
# having weight as wt in K-ary tree
def solve(dp, wt, K, M, used) :
             
    # Return 0 if weight becomes less
    # than zero
    if (wt < 0) :
        return 0
 
    if (wt == 0) :
 
        # Return one only if the
        # current path has included
        # edge weight of atleast M
        if (used) :
            return 1
        return 0
     
    if (dp[wt][used] != -1) :
        return dp[wt][used]
 
    ans = 0
    for i in range(1, K + 1) :
 
        # If the current edge weight
        # is greater than or equal to
        # M, set used as true
        if (i >= M) :
            ans += solve(dp, wt - i,
                         K, M, used | 1)
        else :
            ans += solve(dp, wt - i,
                         K, M, used)
     
    dp[wt][used] = ans
     
    return ans
 
# Driver Code
if __name__ == "__main__" :
 
    W = 3
    K = 3
    M = 2
    dp = np.ones((W + 1, 2));
    dp = -1 * dp
    print(solve(dp, W, K, M, 0))
 
# This code is contributed by Ryuga

// C# program to count the number of
// paths with weight W in a K-ary tree
using System;
   
class GFG
{
    // Function to return the number of ways
    // having weight as wt in K-ary tree
    public static int solve(int[,] dp, int wt, int K, int M, int used)
    {
        // Return 0 if weight becomes less
        // than zero
        if (wt < 0)
            return 0;
       
        if (wt == 0) {
       
            // Return one only if the
            // current path has included
            // edge weight of atleast M
            if (used == 1)
                return 1;
            return 0;
        }
       
        if (dp[wt,used] != -1)
            return dp[wt,used];
       
        int ans = 0;
        for (int i = 1; i <= K; i++) {
       
            // If the current edge weight
            // is greater than or equal to
            // M, set used as true
            if (i >= M)
                ans += solve(dp, wt - i,
                             K, M, used | 1);
            else
                ans += solve(dp, wt - i,
                             K, M, used);
        }
        return dp[wt,used] = ans;
    }
       
    // Driver Code to test above function
    static void Main()
    {
        int W = 3, K = 3, M = 2;
        int[,] dp = new int[W + 1,2];
        for(int i = 0;i < W + 1; i++)
            for(int j = 0; j < 2; j++)
                dp[i,j] = -1;
        Console.Write(solve(dp, W, K, M, 0) + "\n");
    }
    //This code is contributed by DrRoot_
}

<script>
    // Javascript program to count the number of
    // paths with weight W in a K-ary tree
     
    // Function to return the number of ways
    // having weight as wt in K-ary tree
   
    function solve(dp, wt, K, M, used)
    {
        // Return 0 if weight becomes less
        // than zero
        if (wt < 0)
        {
            return 0;
        }
   
        if (wt == 0)
        {
   
            // Return one only if the
            // current path has included
            // edge weight of atleast M
            if (used == 1)
            {
                return 1;
            }
            return 0;
        }
   
        if (dp[wt][used] != -1)
        {
            return dp[wt][used];
        }
   
        let ans = 0;
        for (let i = 1; i <= K; i++)
        {
   
            // If the current edge weight
            // is greater than or equal to
            // M, set used as true
            if (i >= M)
            {
                ans += solve(dp, wt - i,
                        K, M, used | 1);
            }
            else
            {
                ans += solve(dp, wt - i,
                        K, M, used);
            }
        }
        return dp[wt][used] = ans;
    }
     
    let W = 3, K = 3, M = 2;
    let dp = new Array(W + 1);
    for (let i = 0; i < W + 1; i++)
    {
        dp[i] = new Array(2);
      for (let j = 0; j < 2; j++)
      {
        dp[i][j] = -1;
      }
    }
    document.write(solve(dp, W, K, M, 0) + "</br>");
    
   // This code is contributed by suresh07.
</script>

3

Complejidad de tiempo: O(W * K)