查找度数序列是否可以形成一个简单的图形| Havel-Hakimi算法

给定非负整数arr []的序列，任务是检查是否存在与该度序列相对应的简单图。请注意，简单图是没有自环和平行边的图。

例子：

Input: arr[] = {3, 3, 3, 3} 
Output: Yes 
This is actually a complete graph(K₄)

Input: arr[] = {3, 2, 1, 0} 
Output: No 
A vertex has degree n-1 so it’s connected to all the other n-1 vertices. 
But another vertex has degree 0 i.e. isolated. It’s a contradiction. 

方法：一种检查简单图形是否存在的方法是使用Havel-Hakimi算法，如下所示：

• 以非递增顺序对非负整数序列进行排序。
• 删除第一个元素(例如V)。从下一个V元素中减去1。
• 重复1和2，直到满足其中一个停止条件。

停止条件：

• 剩余的所有元素都等于0(存在简单图)。
• 减后遇到负数(不存在简单图形)。
• 没有足够的元素可用于减法步骤(不存在简单图形)。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if
// a simple graph exists
bool graphExists(vector &a, int n)
{
    // Keep performing the operations until one
    // of the stopping condition is met
    while (1)
    {
        // Sort the list in non-decreasing order
        sort(a.begin(), a.end(), greater<>());
 
        // Check if all the elements are equal to 0
        if (a[0] == 0)
            return true;
 
        // Store the first element in a variable
        // and delete it from the list
        int v = a[0];
        a.erase(a.begin() + 0);
 
        // Check if enough elements
        // are present in the list
        if (v > a.size())
            return false;
 
        // Subtract first element from next v elements
        for (int i = 0; i < v; i++)
        {
            a[i]--;
 
            // Check if negative element is
            // encountered after subtraction
            if (a[i] < 0)
                return false;
        }
    }
}
 
// Driver Code
int main()
{
    vector a = {3, 3, 3, 3};
    int n = a.size();
 
    graphExists(a, n) ? cout << "Yes" :
                        cout << "NO" << endl;
 
    return 0;
}
 
// This code is contributed by
// sanjeev2552

Java

// Java implementation of the approach
import java.util.*;
 
@SuppressWarnings("unchecked")
 
class GFG{
   
// Function that returns true if
// a simple graph exists
static boolean graphExists(ArrayList a, int n)
{
     
    // Keep performing the operations until one
    // of the stopping condition is met
    while (true)
    {
         
        // Sort the list in non-decreasing order
        Collections.sort(a, Collections.reverseOrder());
          
        // Check if all the elements are equal to 0
        if ((int)a.get(0) == 0)
            return true;
    
        // Store the first element in a variable
        // and delete it from the list
        int v = (int)a.get(0);
        a.remove(a.get(0));
    
        // Check if enough elements
        // are present in the list
        if (v > a.size())
            return false;
    
        // Subtract first element from
        // next v elements
        for(int i = 0; i < v; i++)
        {
            a.set(i, (int)a.get(i) - 1);
    
            // Check if negative element is
            // encountered after subtraction
            if ((int)a.get(i) < 0)
                return false;
        }
    }
}
   
// Driver Code
public static void main(String[] args)
{
    ArrayList a = new ArrayList();
    a.add(3);
    a.add(3);
    a.add(3);
    a.add(3);
     
    int n = a.size();
      
    if (graphExists(a, n))
    {
        System.out.print("Yes");
    }
    else
    {
        System.out.print("NO");
    }
}
}
 
// This code is contributed by pratham76

Python3

# Python3 implementation of the approach
 
# Function that returns true if
# a simple graph exists
def graphExists(a):
 
    # Keep performing the operations until one
    # of the stopping condition is met
    while True:
 
        # Sort the list in non-decreasing order
        a = sorted(a, reverse = True)
 
        # Check if all the elements are equal to 0
        if a[0]== 0 and a[len(a)-1]== 0:
            return True
 
        # Store the first element in a variable
        # and delete it from the list
        v = a[0]
        a = a[1:]
 
        # Check if enough elements
        # are present in the list
        if v>len(a):
            return False
 
        # Subtract first element from next v elements
        for i in range(v):
            a[i]-= 1
 
            # Check if negative element is
            # encountered after subtraction
            if a[i]<0:
                return False
 
 
# Driver code
a = [3, 3, 3, 3]
if(graphExists(a)):
    print("Yes")
else:
    print("No")

C#

// C# implementation of the approach
using System;
using System.Collections;
 
class GFG{
  
// Function that returns true if
// a simple graph exists
static bool graphExists(ArrayList a, int n)
{
     
    // Keep performing the operations until one
    // of the stopping condition is met
    while (true)
    {
         
        // Sort the list in non-decreasing order
        a.Sort();
        a.Reverse();
         
        // Check if all the elements are equal to 0
        if ((int)a[0] == 0)
            return true;
   
        // Store the first element in a variable
        // and delete it from the list
        int v = (int)a[0];
        a.Remove(a[0]);
   
        // Check if enough elements
        // are present in the list
        if (v > a.Count)
            return false;
   
        // Subtract first element from
        // next v elements
        for(int i = 0; i < v; i++)
        {
            a[i] = (int)a[i] - 1;
   
            // Check if negative element is
            // encountered after subtraction
            if ((int)a[i] < 0)
                return false;
        }
    }
}
  
// Driver Code
public static void Main(string[] args)
{
    ArrayList a = new ArrayList(){ 3, 3, 3, 3 };
    int n = a.Count;
     
    if (graphExists(a, n))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("NO");
    }
}
}
 
// This code is contributed by rutvik_56

输出：

Yes