查找无向图是否包含给定大小的独立集合

给定一个无向图，请检查它是否包含大小为k的独立集合。如果存在独立的大小为k的集合，则打印“是”。否则打印“否”。

独立集：图形中的独立集是一组不相互直接连接的顶点。

范例1：

Input : K = 4,
        graph = [[1, 1, 0, 0, 0],
                 [1, 1, 1, 1, 1],
                 [0, 1, 1, 0, 0],
                 [0, 1, 0, 1, 0],
                 [0, 1, 0, 0, 1]]
Output : Yes 

上图包含一组独立的大小4(顶点1、2、3、4彼此不直接连接)。因此，输出为“是”。

范例2：

Input : K = 4,
        graph = [[1, 1, 0, 0, 0],
                 [1, 1, 1, 1, 1],
                 [0, 1, 1, 0, 0],
                 [0, 1, 0, 1, 1],
                 [0, 1, 0, 1, 1]]
Output : No

上图不包含大小为4的独立集合。因此，输出为“否”。

方法：

• 使用布尔False值初始化变量sol 。
• 从给定图中找到大小为K的所有可能的顶点集。
• 如果找到独立的大小为k的集合，请将sol的值更改为True并返回。
• 否则，继续检查其他可能的设置。
• 最后，如果sol为True，则打印“是”，否则打印“否”。

下面是上述方法的实现：

C++14

// C++ code to check if a given graph
// contains an independent set of size k
#include 
using namespace std;
 
// Function prototype
bool check(int[][5], vector &, int);
 
// Function to construct a set of given size k
bool func(int graph[][5], vector &arr,
          int k, int index, bool sol[])
{
    // Check if the selected set is independent or not.
    // Change the value of sol to True and return
    // if it is independent
    if (k == 0)
    {
        if (check(graph, arr, arr.size()))
        {
            sol[0] = true;
            return true;
        }
    }
    else
    {
        // Set of size k can be formed even if we don't
        // include the vertex at current index.
        if (index >= k)
        {
            vector newvec(arr.begin(), arr.end());
            newvec.push_back(index);
            return (func(graph, newvec, k - 1,
                                index - 1, sol) or
                    func(graph, arr, k, index - 1, sol));
        }
 
        // Set of size k cannot be formed if we don't
        // include the vertex at current index.
        else
        {
            arr.push_back(index);
            return func(graph, arr, k - 1,
                          index - 1, sol);
        }
    }
}
 
// Function to check if the given set is
// independent or not
// arr --> set of size k (contains the
// index of included vertex)
bool check(int graph[][5], vector &arr, int n)
{
    // Check if each vertex is connected to any other
    // vertex in the set or not
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (graph[arr[i]][arr[j]] == 1)
                return false;
    return true;
}
 
// Driver Code
int main()
{
    int graph[][5] = {{1, 1, 0, 0, 0},
                      {1, 1, 1, 1, 1},
                      {0, 1, 1, 0, 0},
                      {0, 1, 0, 1, 0},
                      {0, 1, 0, 0, 1}};
    int k = 4;
    vector arr; // Empty set
    bool sol[] = {false};
    int n = sizeof(graph) /
            sizeof(graph[0]);
    func(graph, arr, k, n - 1, sol);
 
    if (sol[0])
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}
 
// This code is contributed by
// sanjeev2552

Java

// Java code to check if a
// given graph contains an
// independent set of size k
import java.util.*;
class GFG{
 
static  Vector arr =
        new Vector<>();
 
// Function to cona set of
// given size k
static boolean func(int graph[][],
                    int k, int index,
                    boolean sol[])
{
  // Check if the selected set is
  // independent or not. Change the
  // value of sol to True and return
  // if it is independent
  if (k == 0)
  {
    if (check(graph,
              arr.size()))
    {
      sol[0] = true;
      return true;
    }
  }
  else
  {
    // Set of size k can be
    // formed even if we don't
    // include the vertex at
    // current index.
    if (index >= k)
    {
      Vector newvec =
             new Vector<>();
      newvec.add(index);
      return (func(graph, k - 1,
                   index - 1, sol) ||
              func(graph, k,
                   index - 1, sol));
    }
 
    // Set of size k cannot be
    // formed if we don't include
    // the vertex at current index.
    else
    {
      arr.add(index);
      return func(graph, k - 1,
                  index - 1, sol);
    }
  }
  return true;
}
 
// Function to check if the
// given set is independent
// or not arr -. set of size
// k (contains the index of
// included vertex)
static boolean check(int graph[][],
                     int n)
{
  // Check if each vertex is
  // connected to any other
  // vertex in the set or not
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (graph[arr.get(i)][arr.get(j)] == 1)
        return false;
  return true;
}
 
// Driver Code
public static void main(String[] args)
{
  int graph[][] = {{1, 1, 0, 0, 0},
                   {1, 1, 1, 1, 1},
                   {0, 1, 1, 0, 0},
                   {0, 1, 0, 1, 0},
                   {0, 1, 0, 0, 1}};
  int k = 4;
  boolean []sol = new boolean[1];
  int n = graph.length;
  func(graph, k, n - 1, sol);
 
  if (sol[0])
    System.out.print("Yes" + "\n");
  else
    System.out.print("No" + "\n");
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 code to check if a given graph
# contains an independent set of size k
 
# Function to construct a set of given size k
def func(graph, arr, k, index, sol):
     
    # Check if the selected set is independent or not.
    # Change the value of sol to True and return
    # if it is independent
    if k == 0:
        if check(graph, arr) == True:
            sol[0] = True
            return
         
    else:
        # Set of size k can be formed even if we don't
        # include the vertex at current index.
        if index >= k:
            return (func(graph, arr[:] + [index], k-1, index-1, sol) or
                    func(graph, arr[:], k, index-1, sol))
         
        # Set of size k cannot be formed if we don't
        # include the vertex at current index.
        else:
            return func(graph, arr[:] + [index], k-1, index-1, sol)
 
# Function to check if the given set is
# independent or not
# arr --> set of size k (contains the
# index of included vertex)
def check(graph, arr):
     
    # Check if each vertex is connected to any other
    # vertex in the set or not
    for i in range(len(arr)):
        for j in range(i + 1, len(arr)):
             
            if graph[arr[i]][arr[j]] == 1:
                return False
                 
    return True
 
# Driver Code   
graph = [[1, 1, 0, 0, 0],
         [1, 1, 1, 1, 1],
         [0, 1, 1, 0, 0],
         [0, 1, 0, 1, 0],
         [0, 1, 0, 0, 1]]
         
k = 4
arr = []     # Empty set
sol = [False]
 
func(graph, arr[:], k, len(graph)-1, sol)
 
if sol[0] == True:
    print("Yes")
else:
    print("No")

C#

// C# code to check if a
// given graph contains an
// independent set of size k
using System;
using System.Collections.Generic;
 
class GFG{
 
static List arr = new List();
 
// Function to cona set of
// given size k
static bool func(int [,]graph,
                 int k, int index,
                 bool []sol)
{
   
  // Check if the selected set is
  // independent or not. Change the
  // value of sol to True and return
  // if it is independent
  if (k == 0)
  {
    if (check(graph,
              arr.Count))
    {
      sol[0] = true;
      return true;
    }
  }
  else
  {
     
    // Set of size k can be
    // formed even if we don't
    // include the vertex at
    // current index.
    if (index >= k)
    {
      List newvec = new List();
      newvec.Add(index);
       
      return (func(graph, k - 1,
                       index - 1, sol) ||
              func(graph, k,
                   index - 1, sol));
    }
     
    // Set of size k cannot be
    // formed if we don't include
    // the vertex at current index.
    else
    {
      arr.Add(index);
       
      return func(graph, k - 1,
                     index - 1, sol);
    }
  }
  return true;
}
 
// Function to check if the
// given set is independent
// or not arr -. set of size
// k (contains the index of
// included vertex)
static bool check(int [,]graph, int n)
{
   
  // Check if each vertex is
  // connected to any other
  // vertex in the set or not
  for(int i = 0; i < n; i++)
    for(int j = i + 1; j < n; j++)
      if (graph[arr[i],arr[j]] == 1)
        return false;
   
  return true;
}
 
// Driver Code
public static void Main(String[] args)
{
  int [,]graph = { { 1, 1, 0, 0, 0 },
                   { 1, 1, 1, 1, 1 },
                   { 0, 1, 1, 0, 0 },
                   { 0, 1, 0, 1, 0 },
                   { 0, 1, 0, 0, 1 } };
  int k = 4;
  bool []sol = new bool[1];
  int n = graph.GetLength(0);
   
  func(graph, k, n - 1, sol);
 
  if (sol[0])
    Console.Write("Yes" + "\n");
  else
    Console.Write("No" + "\n");
}
}
 
// This code is contributed by Amit Katiyar

输出：

Yes

时间复杂度：

$O({V\choose k} * k^2)$

其中V是图形中的顶点数， k是给定的set大小。