Find the number of distinct pairs of vertices which have a distance of exactly k in a tree

Given an integer k and a tree with n nodes. The task is to count the number of distinct pairs of vertices which have a distance of exactly k.

Examples:

Input: k = 2

Output: 4

Input: k = 3

Output: 2

Approach: This problem can be solved using dynamic programming. For every vertex v of the tree, we calculate values d[v][lev] (0 <= lev <= k). This value indicates the number of vertices having distance lev from v. Note that d[v][0] = 0.
Then we calculate the answer. For any vertex v number of pairs will be a product of the number of vertices at level j – 1 and level k – j.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 5005
  
// To store vertices and value of k
int n, k;
  
vector<int> 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;
}

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Output:

4


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