使用 N 个顶点构建图，其中 K 对顶点之间的最短距离为 2

给定两个正整数N和K ，任务是构造一个由N个顶点组成的简单连通图，其中每条边的长度为1 个单位，使得恰好K对顶点之间的最短距离为2 。如果无法构建图形，则打印-1 。否则，打印图形的边缘。

例子：

Input: N = 5, K = 3
Output: { { 1, 2 }, { 1, 3}, { 1, 4 }, { 1, 5 }, { 2, 3 }, { 2, 4 }, { 2, 5 } }
Explanation:
The distance between the pairs of vertices { (3, 4), (4, 5), (3, 5) } is 2.

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

编程需要懂一点英语

处理方法：按照以下步骤解决问题：

由于图简单且连通，因此，最大可能的边数，比如说Max是((N – 1) * (N – 2)) / 2 。
如果K大于Max ，则打印-1 。
初始化一个数组，比如edges[] ，来存储图的边。
否则，首先将所有顶点与1连接并将其存储在edge[] 中，然后连接所有顶点对(i, j)使得i >= 2和j > i并将其存储在edge[] 中。
最后，打印edge[]数组的第一个((N – 1) + Max – K ) 个元素。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
#include 
using namespace std;
 
// Function to construct the simple and
// connected graph such that the distance
// between exactly K pairs of vertices is 2
void constGraphWithCon(int N, int K)
{
 
    // Stores maximum possible count
    // of edges in a graph
    int Max = ((N - 1) * (N - 2)) / 2;
 
    // Base Case
    if (K > Max) {
        cout << -1 << endl;
        return;
    }
 
    // Stores edges of a graph
    vector > ans;
 
    // Connect all vertices of pairs (i, j)
    for (int i = 1; i < N; i++) {
        for (int j = i + 1; j <= N; j++) {
            ans.emplace_back(make_pair(i, j));
        }
    }
 
    // Print first ((N - 1) + Max - K)  elements
    // of edges[]
    for (int i = 0; i < (N - 1) + Max - K; i++) {
        cout << ans[i].first << " "
             << ans[i].second << endl;
    }
}
 
// Driver Code
int main()
{
    int N = 5, K = 3;
    constGraphWithCon(N, K);
 
    return 0;
}

C

// C program to implement
// the above approach
 
#include 
 
// Function to construct the simple and
// connected graph such that the distance
// between exactly K pairs of vertices is 2
void constGraphWithCon(int N, int K)
{
 
    // Stores maximum possible count
    // of edges in a graph
    int Max = ((N - 1) * (N - 2)) / 2;
 
    // Base Case
    if (K > Max) {
        printf("-1");
        return;
    }
 
    // Stores count of edges in a graph
    int count = 0;
 
    // Connect all vertices of pairs (i, j)
    for (int i = 1; i < N; i++) {
 
        for (int j = i + 1; j <= N; j++) {
 
            printf("%d %d\n", i, j);
 
            // Update
            count++;
 
            if (count == N * (N - 1) / 2 - K)
                break;
        }
 
        if (count == N * (N - 1) / 2 - K)
            break;
    }
}
 
// Driver Code
int main()
{
    int N = 5, K = 3;
    constGraphWithCon(N, K);
 
    return 0;
}

Java

// Java program to implement
// the above 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 construct the simple and connected
// graph such that the distance between
// exactly K pairs of vertices is 2
static void constGraphWithCon(int N, int K)
{
     
    // Stores maximum possible count
    // of edges in a graph
    int Max = ((N - 1) * (N - 2)) / 2;
 
    // Base Case
    if (K > Max)
    {
        System.out.print(-1 + "\n");
        return;
    }
 
    // Stores edges of a graph
    Vector ans = new Vector<>();
 
    // Connect all vertices of pairs (i, j)
    for(int i = 1; i < N; i++)
    {
        for(int j = i + 1; j <= N; j++)
        {
            ans.add(new pair(i, j));
        }
    }
 
    // Print first ((N - 1) + Max - K)  elements
    // of edges[]
    for(int i = 0; i < (N - 1) + Max - K; i++)
    {
        System.out.print(ans.get(i).first + " " +
                         ans.get(i).second +"\n");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 5, K = 3;
     
    constGraphWithCon(N, K);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
# Function to construct the simple and
# connected graph such that the distance
# between exactly K pairs of vertices is 2
def constGraphWithCon(N, K):
   
    # Stores maximum possible count
    # of edges in a graph
    Max = ((N - 1) * (N - 2)) // 2
     
    # Base case
    if (K > Max):
        print(-1)
        return
     
    # Stores edges of a graph
    ans = []
 
    # Connect all vertices of pairs (i, j)
    for i in range(1, N):
        for j in range(i + 1, N + 1):
            ans.append([i, j])
             
    # Print first ((N - 1) + Max - K)  elements
    # of edges[]
    for i in range(0, (N - 1) + Max - K):
        print(ans[i][0], ans[i][1], sep = " ")
 
# Driver code
if __name__ == '__main__':
     
    N = 5
    K = 3
     
    constGraphWithCon(N, K)
 
# This code is contributed by MuskanKalra1

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
     
class pair
{
    public int first, second;
     
    public pair(int first, int second) 
    {
        this.first = first;
        this.second = second;
    }   
}
 
// Function to construct the simple and connected
// graph such that the distance between
// exactly K pairs of vertices is 2
static void constGraphWithCon(int N, int K)
{
     
    // Stores maximum possible count
    // of edges in a graph
    int Max = ((N - 1) * (N - 2)) / 2;
 
    // Base Case
    if (K > Max)
    {
        Console.Write(-1 + "\n");
        return;
    }
 
    // Stores edges of a graph
    List ans = new List();
 
    // Connect all vertices of pairs (i, j)
    for(int i = 1; i < N; i++)
    {
        for(int j = i + 1; j <= N; j++)
        {
            ans.Add(new pair(i, j));
        }
    }
 
    // Print first ((N - 1) + Max - K)  elements
    // of edges[]
    for(int i = 0; i < (N - 1) + Max - K; i++)
    {
        Console.Write(ans[i].first + " " +
                         ans[i].second +"\n");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 5, K = 3;
    constGraphWithCon(N, K);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

输出：

时间复杂度： O(N ² )
辅助空间： O(N ² )

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