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📜  检查在满足给定条件的图中是否存在长度为3的循环

📅  最后修改于: 2021-04-22 09:57:11             🧑  作者: Mango

给定一个代表图形节点的N个整数的Arr数组。边缘被一对那些其位与不等于之间。任务是查找图中是否存在长度为3的循环。

例子:

天真的方法:
运行一个嵌套循环,并检查每对之间是否存在边(如果它们的按位AND值不为零)。因此,我们形成了整个图,并使用任何周期检测方法来检查该图中是否存在周期。

高效方法:

  • 通过连接至少3个边来形成一个循环。
  • 想法是制作一个二维数组,其中第i行存储Arr [i]的二进制值。
  • 接下来,以列方式循环并检查是否存在1的数目至少为3的列。
    • 如果是这样,则表明图中存在一个循环。
    • 如果不存在这样的列,则表示图中没有循环。

下面是上述方法的实现:

C++
// C++ implementation of
// the above approach
#include 
using namespace std;
  
// Maximum numner of bits in Arr[i]
#define MAX 64
  
// Function to check if a cycle
// exists in the given graph
void checkCycle(int ar[], int n)
{
    int bi[n][MAX] = { 0 }, size;
    bool flag = false;
  
    // storing the given ar[i] in
    // binary in the array
    // where bi[i] is the
    // binary value of ar[i]
    for (int i = 0; i < n; i++) {
        size = 0;
        while (ar[i] != 0) {
            bi[i][size++] = ar[i] % 2;
            ar[i] = ar[i] / 2;
        }
    }
  
    // Checking if any column
    // contains at least 3 1's
    for (int i = 0; i < MAX; i++) {
        int ctr = 0;
        for (int j = 0; j < n; j++) {
            if (bi[j][i] == 1)
                ctr = ctr + 1;
        }
        if (ctr >= 3) {
  
            // If number of 1's is more than
            // 3, set flag to true and break
            flag = true;
            break;
        }
    }
  
    // if flag is true, it implies
    // that graph contains a cycle
    if (flag)
        cout << "Yes";
    // if flag is false, no cycle
    // is there in the graph
    else
        cout << "No";
}
  
// Driver code
int main()
{
    int ar[] = { 26, 33, 35, 40, 50 };
    int n = sizeof(ar) / sizeof(ar[0]);
    checkCycle(ar, n);
}


Java
// Java implementation of
// the above approach
import java.io.*;
  
class GFG 
{
      
// Maximum numner of bits in Arr[i]
static int MAX = 64;
  
// Function to check if a cycle
// exists in the given graph
static void checkCycle(int ar[], int n)
{
    int bi[][] = new int[n][MAX];
    int size;
    boolean flag = false;
  
    // storing the given ar[i] in
    // binary in the array
    // where bi[i] is the
    // binary value of ar[i]
    for (int i = 0; i < n; i++) 
    {
        size = 0;
        while (ar[i] != 0) 
        {
            bi[i][size++] = ar[i] % 2;
            ar[i] = ar[i] / 2;
        }
    }
  
    // Checking if any column
    // contains at least 3 1's
    for (int i = 0; i < MAX; i++)
    {
        int ctr = 0;
        for (int j = 0; j < n; j++) 
        {
            if (bi[j][i] == 1)
                ctr = ctr + 1;
        }
        if (ctr >= 3) 
        {
  
            // If number of 1's is more than
            // 3, set flag to true and break
            flag = true;
            break;
        }
    }
  
    // if flag is true, it implies
    // that graph contains a cycle
    if (flag)
        System.out.println ("Yes");
          
    // if flag is false, no cycle
    // is there in the graph
    else
        System.out.println("No");
}
  
// Driver code
public static void main (String[] args)
{
    int ar[] = { 26, 33, 35, 40, 50 };
    int n = ar.length;
    checkCycle(ar, n);
}
}
  
// The code is contrubted by ajit


Python3
# Python3 implementation of
# the above approach
  
# Maximum numner of bits in Arr[i]
MAX = 64
  
# Function to check if a cycle
# exists in the given graph
def checkCycle(ar, n):
    bi = [[0 for i in range(MAX)] 
             for j in range(n)]
    size = -1
    flag = False
  
    # storing the given ar[i] in
    # binary in the array
    # where bi[i] is the
    # binary value of ar[i]
    for i in range(n):
        size = 0
        while ar[i]:
            bi[i][size] = ar[i] % 2
            ar[i] //= 2
            size += 1
  
    # Checking if any column
    # contains at least 3 1's
    for i in range(MAX):
        ctr = 0
        for j in range(n):
            if bi[j][i] == 1:
                ctr += 1
  
        if ctr >= 3:
  
            # If number of 1's is more than
            # 3, set flag to true and break
            flag = True
            break
  
    # if flag is true, it implies
    # that graph contains a cycle
    if flag:
        print("Yes")
  
    # if flag is false, no cycle
    # is there in the graph
    else:
        print("No")
  
# Driver Code
if __name__ == "__main__":
    ar = [26, 33, 35, 40, 50]
    n = len(ar)
    checkCycle(ar, n)
      
# This code is contributed by
# sanjeev2552


C#
// C# implementation of
// the above approach
using System;
  
class GFG 
{
      
// Maximum numner of bits in Arr[i]
static int MAX = 64;
  
// Function to check if a cycle
// exists in the given graph
static void checkCycle(int []ar, int n)
{
    int [,]bi = new int[n, MAX];
    int size;
    Boolean flag = false;
  
    // storing the given ar[i] in
    // binary in the array
    // where bi[i] is the
    // binary value of ar[i]
    for (int i = 0; i < n; i++) 
    {
        size = 0;
        while (ar[i] != 0) 
        {
            bi[i, size++] = ar[i] % 2;
            ar[i] = ar[i] / 2;
        }
    }
  
    // Checking if any column
    // contains at least 3 1's
    for (int i = 0; i < MAX; i++)
    {
        int ctr = 0;
        for (int j = 0; j < n; j++) 
        {
            if (bi[j, i] == 1)
                ctr = ctr + 1;
        }
        if (ctr >= 3) 
        {
  
            // If number of 1's is more than
            // 3, set flag to true and break
            flag = true;
            break;
        }
    }
  
    // if flag is true, it implies
    // that graph contains a cycle
    if (flag)
        Console.WriteLine("Yes");
          
    // if flag is false, no cycle
    // is there in the graph
    else
        Console.WriteLine("No");
}
  
// Driver code
public static void Main (String[] args)
{
    int []ar = { 26, 33, 35, 40, 50 };
    int n = ar.Length;
    checkCycle(ar, n);
}
}
  
// This code is contributed by PrinciRaj1992


输出:
Yes