K元树中权重W的路径数

给定一个K进制树，其中每个节点都有K个子节点，每个边都有一定权重。从特定节点到其所有子节点的所有边缘(即K)的权重按升序1、2、3，…，K求。总权重为W的路径数(路径中所有边缘权重的总和) )从根开始，并至少包含一个重量至少为M的边。

例子：

输入： W = 3，K = 3，M = 2输出： 3说明：一条路径可以是(1 + 2)，第二条路径可以是(2 + 1)，第三条路径可以是3。 输入： W = 4，K = 3，M = 2输出： 6

方法：可以使用动态编程方法解决此问题。这个想法是要保持两种状态，一种状态是当前重量的要求，另一种状态是布尔变量的，这表明当前路径是否包括了一条至少有权重M的边缘。迭代所有可能的边缘权重，即K，并递归求解权重W – i，满足1≤i≤K的要求。如果当前边缘权重大于或等于M，则将下一个调用的布尔变量设置为1。

下面是上述方法的实现。

C++

// C++ program to count the number of
// paths with weight W in a K-ary tree
#include 
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

// 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

Python3

# 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#

// 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_
}

输出：

3

时间复杂度： O(W * K)