给定直径、高度和顶点的树的可能边

找到具有给定值的树并打印树的边缘。如果树是不可能的，打印“-1”。

给定三个整数 n、d 和 h。

n -> Number of vertices. [1, n]
d -> Diameter of the tree (largest 
   distance between two vertices).
h -> Height of the tree (longest distance
   between vertex 1 and another vertex)

例子 ：

Input : n = 5, d = 3, h = 2 
Output : 1 2
         2 3
         1 4
         1 5
Explanation :  

We can see that the height of the tree is 2 (1 -> 
2 --> 5) and diameter is 3 ( 3 -> 2 -> 1 -> 5).
So our conditions are satisfied.

Input :  n = 8, d = 4, h = 2
Output : 1 2
         2 3
         1 4
         4 5
         1 6
         1 7
         1 8
Explanation :

1. 观察当d = 1 时，我们不能构造一棵树（如果树有超过 2 个顶点）。同样当d > 2*h 时，我们不能构造一棵树。
2. 正如我们所知，高度是从顶点 1 到另一个顶点的最长路径。因此，通过将边添加到 h 来构建从顶点 1 开始的路径。现在，如果d > h ，我们应该添加另一条路径以满足从顶点 1 开始的直径，长度为d – h 。
3. 我们的高度和直径条件都满足了。但是仍然可能会留下一些顶点。在端点以外的任何顶点添加剩余的顶点。这一步不会改变我们的直径和高度。选择顶点 1 以添加剩余的顶点（您可以选择任何顶点）。
4. 但是当d == h 时，选择顶点 2 来添加剩余的顶点。

C++

// C++ program to construct tree for given count
// width and height.
#include 
using namespace std;
 
// Function to construct the tree
void constructTree(int n, int d, int h)
{
    if (d == 1) {
 
        // Special case when d == 2, only one edge
        if (n == 2 && h == 1) {
            cout << "1 2" << endl;
            return;
        }
        cout << "-1" << endl; // Tree is not possible
        return;
    }
 
    if (d > 2 * h) {
        cout << "-1" << endl;
        return;
    }
 
    // Satisfy the height condition by add
    // edges up to h
    for (int i = 1; i <= h; i++)    
        cout << i << " " << i + 1 << endl;
     
    if (d > h) {
 
        // Add d - h edges from 1 to
        // satisfy diameter condition
        cout << "1"
            << " " << h + 2 << endl;
        for (int i = h + 2; i <= d; i++) {
            cout << i << " " << i + 1 << endl;
        }
    }
 
    // Remaining edges at vertex 1 or 2(d == h)
    for (int i = d + 1; i < n; i++)
    {
        int k = 1;
        if (d == h)
            k = 2;
        cout << k << " " << i + 1 << endl;
    }
}
 
// Driver Code
int main()
{
    int n = 5, d = 3, h = 2;
    constructTree(n, d, h);
    return 0;
}

Java

// Java program to construct tree for given count
// width and height.
class GfG {
 
// Function to construct the tree
static void constructTree(int n, int d, int h)
{
    if (d == 1) {
 
        // Special case when d == 2, only one edge
        if (n == 2 && h == 1) {
            System.out.println("1 2");
            return;
        }
        System.out.println("-1"); // Tree is not possible
        return;
    }
 
    if (d > 2 * h) {
        System.out.println("-1");
        return;
    }
 
    // Satisfy the height condition by add
    // edges up to h
    for (int i = 1; i <= h; i++)    
        System.out.println(i + " " + (i + 1));
     
    if (d > h) {
 
        // Add d - h edges from 1 to
        // satisfy diameter condition
        System.out.println("1" + " " + (h + 2));
        for (int i = h + 2; i <= d; i++) {
            System.out.println(i + " " + (i + 1));
        }
    }
 
    // Remaining edges at vertex 1 or 2(d == h)
    for (int i = d + 1; i < n; i++)
    {
        int k = 1;
        if (d == h)
            k = 2;
        System.out.println(k + " " + (i + 1));
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 5, d = 3, h = 2;
    constructTree(n, d, h);
}
}

Python3

# Python3 code to construct tree for given count
# width and height.
 
# Function to construct the tree
def constructTree(n, d, h):
    if d == 1:
 
        # Special case when d == 2, only one edge
        if n == 2 and h == 1:
            print("1 2")
            return 0
         
        print("-1")    # Tree is not possible
        return 0
     
    if d > 2 * h:
        print("-1")
        return 0
         
    # Satisfy the height condition by add
    # edges up to h
    for i in range(1, h+1):
        print(i," " , i + 1)
     
    if d > h:
 
        # Add d - h edges from 1 to
        # satisfy diameter condition
        print(1,"  ", h + 2)
        for i in range(h+2, d+1):
            print(i, " " , i + 1)
             
    # Remaining edges at vertex 1 or 2(d == h)
    for i in range(d+1, n):
        k = 1
        if d == h:
            k = 2
        print(k ," " , i + 1)
 
# Driver Code
n = 5
d = 3
h = 2
constructTree(n, d, h)
 
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to construct tree for 
// given count width and height.
using System;
 
class GfG
{
 
    // Function to construct the tree
    static void constructTree(int n, int d, int h)
    {
        if (d == 1)
        {
 
            // Special case when d == 2,
            // only one edge
            if (n == 2 && h == 1)
            {
                Console.WriteLine("1 2");
                return;
            }
             
            // Tree is not possible
            Console.WriteLine("-1");
            return;
        }
 
        if (d > 2 * h)
        {
            Console.WriteLine("-1");
            return;
        }
 
        // Satisfy the height condition
        // by add edges up to h
        for (int i = 1; i <= h; i++)
            Console.WriteLine(i + " " + (i + 1));
 
        if (d > h)
        {
 
            // Add d - h edges from 1 to
            // satisfy diameter condition
            Console.WriteLine("1" + " " + (h + 2));
            for (int i = h + 2; i <= d; i++)
            {
                Console.WriteLine(i + " " + (i + 1));
            }
        }
 
        // Remaining edges at vertex 1 or 2(d == h)
        for (int i = d + 1; i < n; i++)
        {
            int k = 1;
            if (d == h)
                k = 2;
            Console.WriteLine(k + " " + (i + 1));
        }
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 5, d = 3, h = 2;
        constructTree(n, d, h);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program to construct tree for
// given count width and height.
 
// Function to construct the tree
function constructTree(n, d, h)
{
    if (d == 1)
    {
         
        // Special case when d == 2,
        // only one edge
        if (n == 2 && h == 1)
        {
            document.write("1 2", "<br>");
            return;
        }
           
        // Tree is not possible
        document.write("-1", "<br>");
        return;
    }
 
    if (d > 2 * h)
    {
        document.write("-1", "<br>");
        return;
    }
 
    // Satisfy the height condition
    // by add edges up to h
    for(var i = 1; i <= h; i++)
        document.write(i + " " + (i + 1), "<br>");
 
    if (d > h)
    {
         
        // Add d - h edges from 1 to
        // satisfy diameter condition
        document.write("1" + " " + (h + 2), "<br>");
        for(var i = h + 2; i <= d; i++)
        {
            document.write(i + " " + (i + 1), "<br>");
        }
    }
 
    // Remaining edges at vertex 1 or 2(d == h)
    for(var i = d + 1; i < n; i++)
    {
        var k = 1;
        if (d == h)
            k = 2;
             
        document.write(k + " " + (i + 1), "<br>");
    }
}
 
// Driver Code
var n = 5, d = 3, h = 2;
 
constructTree(n, d, h);
 
// This code is contributed by bunnyram19
 

输出 ：

1 2
2 3
1 4
1 5