具有N个节点和K个叶子的树，使得最远叶子之间的距离最小

给定N和K，请打印一棵树，以使该树不超过K个叶节点，并且每隔一个节点至少有两个与之连接的节点。任务是精确地构建N个节点的树，以使最远叶节点之间的距离最小。同时打印最小距离。
注意：可以有多棵树。
例子：

Input: N = 5, K = 3 
Output: Distance = 3 
The tree is: 
1 2 
2 3 
3 4 
3 5 
 

Input: N = 3, K = 2 
Output: Distance = 2 
The tree is: 
1 2 
2 3 

方法：

• 最初，树将有k-1个节点连接到1。
• 然后将一个节点一个接一个地连接到所有k-1个节点。
• 如果剩下节点，请保持将它们一一连接到叶节点。

如何构建树的图形表示将使事情变得更加清晰。在下面的图片中，K = 6，并且对于任何数量的N都得到了证明。黄色的节点是叶节点。

下面是上述方法的实现：

C++

// C++ program of above approach
#include 
using namespace std;
 
// Function to print the distance
// and the tree
void buildTree(int n, int k)
{
    int ans = 2 * ((n - 1) / k) + min((n - 1) % k, 2);
    cout << "Distance = " << ans;
 
    cout << "\nThe tree is:\n";
 
    // print all K-1 leaf nodes attached with 1
    for (int i = 2; i <= k; i++) {
        cout << "1 " << i << endl;
    }
 
    // Join nodes to from other left nodes
    // the last node thus will be the left out leaf node
    for (int i = k + 1; i <= n; i++) {
        cout << i << " " << (i - k) << endl;
    }
}
 
// Driver Code
int main()
{
    int n = 5, k = 3;
 
    buildTree(n, k);
}

Java

// Java program of above approach
import java.util.*;
import java.lang.*;
 
// Function to print the distance
// and the tree
class GFG
{
public void buildTree(int n, int k)
{
    int ans = 2 * ((n - 1) / k) +
            Math.min((n - 1) % k, 2);
    System.out.println("Distance = " + ans);
 
    System.out.println("The tree is: ");
 
    // print all K-1 leaf nodes
    // attached with 1
    for (int i = 2; i <= k; i++)
    {
        System.out.println( "1 " + i );
    }
 
    // Join nodes to from other left
    // nodes the last node thus will
    // be the left out leaf node
    for (int i = k + 1; i <= n; i++)
    {
        System.out.println( i + " " +
                        (i - k));
    }
}
 
// Driver Code
public static void main(String args[])
{
    GFG g = new GFG();
    int n = 5, k = 3;
 
    g.buildTree(n, k);
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Python3

# Python3 program of above approach
 
# Function to print the distance
# and the tree
def buildTree(n, k):
 
    ans = (2 * ((n - 1) // k) +
            min((n - 1) % k, 2))
    print("Distance = ", ans )
 
    print("The tree is:")
 
    # print all K-1 leaf nodes
    # attached with 1
    for i in range(2, k + 1):
        print("1 ", i)
     
    # Join nodes to from other left nodes
    # the last node thus will be the
    # left out leaf node
    for i in range(k + 1, n + 1):
        print(i, "", (i - k))
 
# Driver Code
if __name__ == '__main__':
    n = 5
    k = 3
    buildTree(n, k)
     
# This code is contributed
# by SHUBHAMSINGH10

C#

// C# program of above approach
using System;
 
// Function to print the distance
// and the tree
class GFG
{
public void buildTree(int n, int k)
{
    int ans = 2 * ((n - 1) / k) +
        Math.Min((n - 1) % k, 2);
    Console.WriteLine("Distance = " + ans);
 
    Console.WriteLine ("The tree is: ");
 
    // print all K-1 leaf nodes
    // attached with 1
    for (int i = 2; i <= k; i++)
    {
        Console.WriteLine( "1 " + i );
    }
 
    // Join nodes to from other left
    // nodes the last node thus will
    // be the left out leaf node
    for (int i = k + 1; i <= n; i++)
    {
        Console.WriteLine ( i + " " +
                        (i - k));
    }
}
 
// Driver Code
public static void Main()
{
    GFG g = new GFG();
    int n = 5, k = 3;
 
    g.buildTree(n, k);
}
}
 
// This code is contributed by Soumik

Javascript

// javascript program of above approach
// Function to prvar the distance
// and the tree
 
    function buildTree(n , k)
    {
        var ans = parseInt(2 * ((n - 1) / k) + Math.min((n - 1) % k, 2));
        document.write("Distance = " + ans+ "<br/>");
 
        document.write("The tree is:<br/> ");
 
        // prvar all K-1 leaf nodes
        // attached with 1
        for (i = 2; i <= k; i++) {
            document.write("1 " + i+"<br/>");
        }
 
        // Join nodes to from other left
        // nodes the last node thus will
        // be the left out leaf node
        for (i = k + 1; i <= n; i++) {
            document.write(i + " " + (i - k)+"<br/>");
        }
    }
 
    // Driver Code
        var n = 5, k = 3;
        buildTree(n, k);
 
// This code is contributed by aashish1995.

输出：

Distance = 3
The tree is:
1 2
1 3
4 1
5 2

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