图中最大和最小孤立顶点

给定图的“n”个顶点和“m”条边。找出图中可能的孤立顶点的最小数量和最大数量。
例子：

Input : 4 2
Output : Minimum 0
         Maximum 1

1--2 3--4 <---Minimum -  No isolated vertex
1--2   <--- Maximum - 1 Isolated vertex i.e. 4
   |
   3


Input : 5 2
Output : Minimum 1
         Maximum 2

1--2 3--4 5  <-- Minimum - 1 isolated vertex i.e. 5
1--2   4  5  <-- Maximum - 2 isolated vertex i.e. 4 and 5
   |
   3

1. 对于最少数量的孤立顶点，我们仅通过一条边连接两个顶点。每个顶点应该只连接到另一个顶点，并且每个顶点都应该有一个度数
   因此，如果边的数量是 ‘m’，并且如果 ‘n’ 顶点 <=2 * ‘m’ 边，则没有孤立顶点，如果此条件为假，则有 n-2*m 个孤立顶点。
2. 为了获得最大数量的孤立顶点，我们创建了一个多边形，这样每个顶点都连接到其他顶点，并且每个顶点与每个其他顶点都有一条对角线。因此，n个边多边形的一个顶点到另一个顶点的对角线数为n*(n-3)/2，连接相邻顶点的边数为n。因此，边的总数为 n*(n-1)/2。

下面是上述方法的实现。

C++

// CPP program to find maximum/minimum number
// of isolated vertices.
#include 
using namespace std;
 
// Function to find out the minimum and
// maximum number of isolated vertices
void find(int n, int m)
{
    // Condition to find out minimum number
    // of isolated vertices
    if (n <= 2 * m)
        cout << "Minimum " << 0 << endl;
    else
        cout << "Minimum " << n - 2 * m << endl;
 
    // To find out maximum number of isolated
    // vertices
    // Loop to find out value of number of
    // vertices that are connected
    int i;
    for (i = 1; i <= n; i++) {
        if (i * (i - 1) / 2 >= m)
            break;
    }
    cout << "Maximum " << n - i;
}
 
// Driver Function
int main()
{
    // Number of vertices
    int n = 4;
 
    // Number of edges
    int m = 2;
 
    // Calling the function to maximum and
    // minimum number of isolated vertices
    find(n, m);
    return 0;
}

Java

// Java program to find maximum/minimum number
// of isolated vertices.
 
import java.io.*;
 
class GFG {
  
 
// Function to find out the minimum and
// maximum number of isolated vertices
 static void find(int n, int m)
{
    // Condition to find out minimum number
    // of isolated vertices
    if (n <= 2 * m)
        System.out.println( "Minimum " + 0);
    else
        System.out.println( "Minimum " + (n - 2 * m));
 
    // To find out maximum number of isolated
    // vertices
    // Loop to find out value of number of
    // vertices that are connected
    int i;
    for (i = 1; i <= n; i++) {
        if (i * (i - 1) / 2 >= m)
            break;
    }
    System.out.println( "Maximum " + (n - i));
}
 
// Driver Function
 
    public static void main (String[] args) {
     
    // Number of vertices
    int n = 4;
 
    // Number of edges
    int m = 2;
 
    // Calling the function to maximum and
    // minimum number of isolated vertices
    find(n, m);
    }
}
//This code is contributed by inder_verma.

Python3

# Python3 program to find maximum/minimum
# number of isolated vertices.
 
# Function to find out the minimum and
# maximum number of isolated vertices
def find(n, m) :
 
    # Condition to find out minimum
    # number of isolated vertices
    if (n <= 2 * m):
        print("Minimum ", 0)
    else:
        print("Minimum ", n - 2 * m )
 
    # To find out maximum number of
    # isolated vertices
    # Loop to find out value of number
    # of vertices that are connected
    for i in range(1, n + 1):
        if (i * (i - 1) // 2 >= m):
            break
     
    print("Maximum ", n - i)
     
# Driver Code
if __name__ == '__main__':
     
    # Number of vertices
    n = 4
 
    # Number of edges
    m = 2
 
    # Calling the function to maximum and
    # minimum number of isolated vertices
    find(n, m)
 
# This code is contributed by
# SHUBHAMSINGH10

C#

// C# program to find maximum/
// minimum number of isolated vertices.
using System;
 
class GFG
{
 
// Function to find out the
// minimum and maximum number
// of isolated vertices
static void find(int n, int m)
{
    // Condition to find out minimum
    // number of isolated vertices
    if (n <= 2 * m)
        Console.WriteLine("Minimum " + 0);
    else
        Console.WriteLine("Minimum " +
                         (n - 2 * m));
 
    // To find out maximum number
    // of isolated vertices
    // Loop to find out value of
    // number of vertices that
    // are connected
    int i;
    for (i = 1; i <= n; i++)
    {
        if (i * (i - 1) / 2 >= m)
            break;
    }
    Console.WriteLine("Maximum " + (n - i));
}
 
// Driver Code
public static void Main ()
{
 
    // Number of vertices
    int n = 4;
     
    // Number of edges
    int m = 2;
     
    // Calling the function to 
    // maximum and minimum number
    // of isolated vertices
    find(n, m);
}
}
 
// This code is contributed
// by inder_verma.

PHP

= $m)
            break;
    }
    echo "Maximum " , ($n - $i);
}
 
// Driver Code
 
// Number of vertices
$n = 4;
 
// Number of edges
$m = 2;
 
// Calling the function to
// maximum and minimum number
// of isolated vertices
find($n, $m);
 
// This code is contributed
// by inder_verma
?>

Javascript

    // Javascript program to find maximum/
    // minimum number of isolated vertices.
     
    // Function to find out the
    // minimum and maximum number
    // of isolated vertices
    function find(n, m)
    {
     
        // Condition to find out minimum
        // number of isolated vertices
        if (n <= 2 * m)
            document.write("Minimum " + 0 + "</br>");
        else
            document.write("Minimum " + (n - 2 * m) + "</br>");
 
        // To find out maximum number
        // of isolated vertices
        // Loop to find out value of
        // number of vertices that
        // are connected
        let i;
        for (i = 1; i <= n; i++)
        {
            if (i * parseInt((i - 1) / 2, 10) >= m)
                break;
        }
        document.write("Maximum " + (n - i));
    }
     
    // Number of vertices
    let n = 4;
       
    // Number of edges
    let m = 2;
       
    // Calling the function to 
    // maximum and minimum number
    // of isolated vertices
    find(n, m);
 
// This code is contributed by divyeshrabadiya07.

输出：

Minimum 0
Maximum 1

时间复杂度– O(n)

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