找出一棵树中距离恰好为k的不同对顶点的数量

给定整数k和具有n个节点的树。任务是计算距离恰好为k的不同顶点对的数量。
例子：

Input: k = 2 
 

Output: 4
Input: k = 3 
 

Output: 2  

方法：可以使用动态编程解决此问题。对于树的每个顶点v，我们计算值d [v] [lev](0 <= lev <= k)。此值表示与v的距离为lev的顶点数。请注意，d [v] [0] = 1。
然后我们计算出答案。对于任何顶点v，线对数将是级别j – 1和级别k – j的顶点数的乘积。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define N 5005
 
// To store vertices and value of k
int n, k;
 
vector gr[N];
 
// To store number vertices at a level i
int d[N][505];
 
// To store the final answer
int ans = 0;
 
// Function to add an edge between two nodes
void Add_edge(int x, int y)
{
    gr[x].push_back(y);
    gr[y].push_back(x);
}
 
// Function to find the number of distinct
// pairs of the vertices which have a distance
// of exactly k in a tree
void dfs(int v, int par)
{
    // At level zero vertex itself is counted
    d[v][0] = 1;
    for (auto i : gr[v]) {
        if (i != par) {
            dfs(i, v);
 
            // Count the pair of vertices at
            // distance k
            for (int j = 1; j <= k; j++)
                ans += d[i][j - 1] * d[v][k - j];
 
            // For all levels count vertices
            for (int j = 1; j <= k; j++)
                d[v][j] += d[i][j - 1];
        }
    }
}
 
// Driver code
int main()
{
    n = 5, k = 2;
 
    // Add edges
    Add_edge(1, 2);
    Add_edge(2, 3);
    Add_edge(3, 4);
    Add_edge(2, 5);
 
    // Function call
    dfs(1, 0);
 
    // Required answer
    cout << ans;
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
    static final int N = 5005;
 
    // To store vertices and value of k
    static int n, k;
 
    static Vector[] gr = new Vector[N];
 
    // To store number vertices at a level i
    static int[][] d = new int[N][505];
 
    // To store the final answer
    static int ans = 0;
 
    // Function to add an edge between two nodes
    static void Add_edge(int x, int y)
    {
        gr[x].add(y);
        gr[y].add(x);
    }
 
    // Function to find the number of distinct
    // pairs of the vertices which have a distance
    // of exactly k in a tree
    static void dfs(int v, int par)
    {
        // At level zero vertex itself is counted
        d[v][0] = 1;
        for (Integer i : gr[v])
        {
            if (i != par)
            {
                dfs(i, v);
 
                // Count the pair of vertices at
                // distance k
                for (int j = 1; j <= k; j++)
                    ans += d[i][j - 1] * d[v][k - j];
 
                // For all levels count vertices
                for (int j = 1; j <= k; j++)
                    d[v][j] += d[i][j - 1];
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        n = 5;
        k = 2;
        for (int i = 0; i < N; i++)
            gr[i] = new Vector();
         
        // Add edges
        Add_edge(1, 2);
        Add_edge(2, 3);
        Add_edge(3, 4);
        Add_edge(2, 5);
 
        // Function call
        dfs(1, 0);
 
        // Required answer
        System.out.print(ans);
 
    }
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
N = 5005
 
# To store vertices and value of k
n, k = 0, 0
 
gr = [[] for i in range(N)]
 
# To store number vertices at a level i
d = [[0 for i in range(505)]
        for i in range(N)]
 
# To store the final answer
ans = 0
 
# Function to add an edge between two nodes
def Add_edge(x, y):
    gr[x].append(y)
    gr[y].append(x)
 
# Function to find the number of distinct
# pairs of the vertices which have a distance
# of exactly k in a tree
def dfs(v, par):
    global ans
     
    # At level zero vertex itself is counted
    d[v][0] = 1
    for i in gr[v]:
        if (i != par):
            dfs(i, v)
 
            # Count the pair of vertices at
            # distance k
            for j in range(1, k + 1):
                ans += d[i][j - 1] * d[v][k - j]
 
            # For all levels count vertices
            for j in range(1, k + 1):
                d[v][j] += d[i][j - 1]
 
# Driver code
n = 5
k = 2
 
# Add edges
Add_edge(1, 2)
Add_edge(2, 3)
Add_edge(3, 4)
Add_edge(2, 5)
 
# Function call
dfs(1, 0)
 
# Required answer
print(ans)
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
    static readonly int N = 5005;
 
    // To store vertices and value of k
    static int n, k;
 
    static List[] gr = new List[N];
 
    // To store number vertices at a level i
    static int[,] d = new int[N, 505];
 
    // To store the readonly answer
    static int ans = 0;
 
    // Function to add an edge between two nodes
    static void Add_edge(int x, int y)
    {
        gr[x].Add(y);
        gr[y].Add(x);
    }
 
    // Function to find the number of distinct
    // pairs of the vertices which have a distance
    // of exactly k in a tree
    static void dfs(int v, int par)
    {
        // At level zero vertex itself is counted
        d[v, 0] = 1;
        foreach (int i in gr[v])
        {
            if (i != par)
            {
                dfs(i, v);
 
                // Count the pair of vertices at
                // distance k
                for (int j = 1; j <= k; j++)
                    ans += d[i, j - 1] * d[v, k - j];
 
                // For all levels count vertices
                for (int j = 1; j <= k; j++)
                    d[v, j] += d[i, j - 1];
            }
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        n = 5;
        k = 2;
        for (int i = 0; i < N; i++)
            gr[i] = new List();
         
        // Add edges
        Add_edge(1, 2);
        Add_edge(2, 3);
        Add_edge(3, 4);
        Add_edge(2, 5);
 
        // Function call
        dfs(1, 0);
 
        // Required answer
        Console.Write(ans);
 
    }
}
 
// This code is contributed by Rajput-Ji

输出：

4

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