检查是否存在满足给定条件的连通图

给定两个整数N和K 。任务是找到一个具有N个顶点的连通图，使得恰好有K对(i, j) ，其中它们之间的最短距离为2 。如果不存在这样的图，则打印-1 。

笔记：

第一行输出应该是图中的边数（比如 m），接下来的 m 行应该包含两个数字，代表顶点之间的边。
如果有多个答案，请打印其中任何一个。

例子：

Input: N = 5, K = 3
Output: 7
1 2
1 3
1 4
1 5
3 4
3 5
4 5

Input: N = 5, K = 8
Output: -1

方法：一个有 N 个顶点的连通图至少有N-1 条边。每对的最短距离等于1 。很明显，如果K > N * (N – 1) / 2 – (N – 1) = (N – 1) * (N – 2) / 2 ，显然不存在解。
反之，通过构造满足条件的图，可以证明如果K ≤ (N – 1) * (N – 2) / 2存在解。首先，让我们考虑每个顶点与所有其他顶点连接的图，然后任何两个顶点之间的最短顶点是1 。现在删除任何K 个边，那么就存在正好K 个这样的对。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to find the required graph
void connected_graph(int n, int k)
{
    // If no such graph exists
    if (k > (n - 1) * (n - 2) / 2) {
        cout << -1 << endl;
        return;
    }
  
    // Consider edge between all vertices
    bool isEdge[n][n] = {};
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++)
            isEdge[i][j] = true;
    }
  
    // Remove K vertices
    int cnt = 0;
    for (int i = 1; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (cnt < k) {
                isEdge[i][j] = false;
                cnt++;
            }
        }
    }
  
    // Store all the edges
    vector > vec;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (isEdge[i][j])
                vec.emplace_back(i, j);
        }
    }
  
    // Print all the edges
    cout << vec.size() << endl;
    for (int i = 0; i < vec.size(); i++) {
        cout << vec[i].first + 1 << " "
             << vec[i].second + 1 << endl;
    }
}
  
// Driver code
int main()
{
    int n = 5, k = 3;
  
    // Function call
    connected_graph(n, k);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG
{
static class pair
{ 
    int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
}
  
// Function to find the required graph
static void connected_graph(int n, int k)
{
    // If no such graph exists
    if (k > (n - 1) * (n - 2) / 2) 
    {
        System.out.println(-1);
        return;
    }
  
    // Consider edge between all vertices
    boolean [][]isEdge = new boolean[n][n];
    for (int i = 0; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++)
            isEdge[i][j] = true;
    }
  
    // Remove K vertices
    int cnt = 0;
    for (int i = 1; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++) 
        {
            if (cnt < k)
            {
                isEdge[i][j] = false;
                cnt++;
            }
        }
    }
  
    // Store all the edges
    Vector vec = new Vector<>();
    for (int i = 0; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++)
        {
            if (isEdge[i][j])
                vec.add(new pair(i, j));
        }
    }
  
    // Print all the edges
    System.out.println(vec.size());
    for (int i = 0; i < vec.size(); i++) 
    {
        System.out.println(vec.get(i).first + 1 + 
                    " " + (vec.get(i).second + 1));
    }
}
  
// Driver code
public static void main(String[] args) 
{
    int n = 5, k = 3;
  
    // Function call
    connected_graph(n, k);
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
import numpy as np;
  
# Function to find the required graph 
def connected_graph(n, k) : 
  
    # If no such graph exists 
    if (k > (n - 1) * (n - 2) / 2) :
        print(-1) ; 
        return; 
  
    # Consider edge between all vertices 
    isEdge = np.zeros((n, n)); 
    for i in range(n) :
        for j in range(i + 1, n) :
            isEdge[i][j] = True; 
  
    # Remove K vertices 
    cnt = 0; 
    for i in range(1, n) :
        for j in range(i + 1 , n) :
            if (cnt < k) :
                isEdge[i][j] = False; 
                cnt += 1; 
  
    # Store all the edges 
    vec = []; 
    for i in range(n) : 
        for j in range(i + 1, n) :
            if (isEdge[i][j]) :
                vec.append([i, j]); 
  
    # Print all the edges 
    print(len(vec)); 
    for i in range(len(vec)) :
        print(vec[i][0] + 1, vec[i][1] + 1); 
  
# Driver code 
if __name__ == "__main__" : 
  
    n = 5; k = 3;
  
    # Function call 
    connected_graph(n, k); 
  
# This code is contributed by Ankit Rai

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG
{
public class pair
{ 
    public int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
}
  
// Function to find the required graph
static void connected_graph(int n, int k)
{
    // If no such graph exists
    if (k > (n - 1) * (n - 2) / 2) 
    {
        Console.WriteLine(-1);
        return;
    }
  
    // Consider edge between all vertices
    bool [,]isEdge = new bool[n, n];
    for (int i = 0; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++)
            isEdge[i, j] = true;
    }
  
    // Remove K vertices
    int cnt = 0;
    for (int i = 1; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++) 
        {
            if (cnt < k)
            {
                isEdge[i, j] = false;
                cnt++;
            }
        }
    }
  
    // Store all the edges
    List vec = new List();
    for (int i = 0; i < n; i++) 
    {
        for (int j = i + 1; j < n; j++)
        {
            if (isEdge[i, j])
                vec.Add(new pair(i, j));
        }
    }
  
    // Print all the edges
    Console.WriteLine(vec.Count);
    for (int i = 0; i < vec.Count; i++) 
    {
        Console.WriteLine(vec[i].first + 1 + 
                   " " + (vec[i].second + 1));
    }
}
  
// Driver code
public static void Main(String[] args) 
{
    int n = 5, k = 3;
  
    // Function call
    connected_graph(n, k);
}
}
  
// This code is contributed by 29AjayKumar

输出：

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