给定图形G ，任务是检查它是否表示环形拓扑。
下图显示了一种环形拓扑：
例子：
Input : Graph =
Output : YES
Input : Graph =
Output : NO如果满足以下三个条件，则V顶点的图形表示环形拓扑：
- 顶点数> = 3。
- 所有顶点的度数应为2 。
- 边数=顶点数。
想法是遍历图形并检查其是否满足以上三个条件。如果是，则表示环形拓扑，否则不表示。
下面是上述方法的实现：
C++
// C++ program to check if the given graph
// represents a Ring topology
#include
using namespace std;
// A utility function to add an edge in an
// undirected graph.
void addEdge(vector adj[], int u, int v)
{
adj[u].push_back(v);
adj[v].push_back(u);
}
// A utility function to print the adjacency list
// representation of graph
void printGraph(vector adj[], int V)
{
for (int v = 0; v < V; ++v) {
cout << "\n Adjacency list of vertex "
<< v << "\n head ";
for (auto x : adj[v])
cout << "-> " << x;
printf("\n");
}
}
/* Function to return true if the graph represented
by the adjacency list represents a Ring topology
else return false */
bool checkRingTopologyUtil(vector adj[], int V, int E)
{
// Number of edges should be equal
// to Number of vertices
if (E != V)
return false;
// For a graph to represent a ring topology should have
// greater than 2 nodes
if (V <= 2)
return false;
int* vertexDegree = new int[V + 1];
memset(vertexDegree, 0, sizeof vertexDegree);
// calculate the degree of each vertex
for (int i = 1; i <= V; i++) {
for (auto v : adj[i]) {
vertexDegree[v]++;
}
}
// countDegree2 stores the count of
// the vertices having degree 2
int countDegree2 = 0;
for (int i = 1; i <= V; i++) {
if (vertexDegree[i] == 2) {
countDegree2++;
}
}
// if all three necessary conditions as discussed,
// satisfy return true
if (countDegree2 == V) {
return true;
}
else {
return false;
}
}
// Function to check if the graph represents a Ring topology
void checkRingTopology(vector adj[], int V, int E)
{
bool isRing = checkRingTopologyUtil(adj, V, E);
if (isRing) {
cout << "YES" << endl;
}
else {
cout << "NO" << endl;
}
}
// Driver code
int main()
{
// Graph 1
int V = 6, E = 6;
vector adj1[V + 1];
addEdge(adj1, 1, 2);
addEdge(adj1, 2, 3);
addEdge(adj1, 3, 4);
addEdge(adj1, 4, 5);
addEdge(adj1, 6, 1);
addEdge(adj1, 5, 6);
checkRingTopology(adj1, V, E);
// Graph 2
V = 5, E = 4;
vector adj2[V + 1];
addEdge(adj2, 1, 2);
addEdge(adj2, 1, 3);
addEdge(adj2, 3, 4);
addEdge(adj2, 4, 5);
checkRingTopology(adj2, V, E);
return 0;
} Java
// Java program to check if the given graph
// represents a Ring topology
import java.util.*;
class GFG{
// A utility function to add an edge in an
// undirected graph.
static void addEdge(Vector adj[],
int u, int v)
{
adj[u].add(v);
adj[v].add(u);
}
// A utility function to print the adjacency list
// representation of graph
static void printGraph(Vector adj[], int V)
{
for(int v = 0; v < V; ++v)
{
System.out.print("\n Adjacency list of vertex " +
v + "\n head ");
for(int x : adj[v])
System.out.print(". " + x);
System.out.printf("\n");
}
}
// Function to return true if the graph represented
// by the adjacency list represents a Ring topology
// else return false
static boolean checkRingTopologyUtil(Vector adj[],
int V, int E)
{
// Number of edges should be equal
// to Number of vertices
if (E != V)
return false;
// For a graph to represent a ring
// topology should have greater
// than 2 nodes
if (V <= 2)
return false;
int[] vertexDegree = new int[V + 1];
// Calculate the degree of each vertex
for(int i = 1; i <= V; i++)
{
for(int v : adj[i])
{
vertexDegree[v]++;
}
}
// countDegree2 stores the count of
// the vertices having degree 2
int countDegree2 = 0;
for(int i = 1; i <= V; i++)
{
if (vertexDegree[i] == 2)
{
countDegree2++;
}
}
// If all three necessary conditions
// as discussed, satisfy return true
if (countDegree2 == V)
{
return true;
}
else
{
return false;
}
}
// Function to check if the graph represents
// a Ring topology
static void checkRingTopology(Vector adj[],
int V, int E)
{
boolean isRing = checkRingTopologyUtil(adj, V, E);
if (isRing)
{
System.out.print("YES" + "\n");
}
else
{
System.out.print("NO" + "\n");
}
}
// Driver code
public static void main(String[] args)
{
// Graph 1
int V = 6, E = 6;
@SuppressWarnings("unchecked")
Vector []adj1 = new Vector[V + 1];
for(int i = 0; i < adj1.length; i++)
adj1[i] = new Vector();
addEdge(adj1, 1, 2);
addEdge(adj1, 2, 3);
addEdge(adj1, 3, 4);
addEdge(adj1, 4, 5);
addEdge(adj1, 6, 1);
addEdge(adj1, 5, 6);
checkRingTopology(adj1, V, E);
// Graph 2
V = 5; E = 4;
@SuppressWarnings("unchecked")
Vector []adj2 = new Vector[V + 1];
for(int i = 0; i < adj2.length; i++)
adj2[i] = new Vector();
addEdge(adj2, 1, 2);
addEdge(adj2, 1, 3);
addEdge(adj2, 3, 4);
addEdge(adj2, 4, 5);
checkRingTopology(adj2, V, E);
}
}
// This code is contributed by Amit Katiyar Python3
# Python3 program to check if the given graph
# represents a star topology
# A utility function to add an edge in an
# undirected graph.
def addEdge(adj, u, v):
adj[u].append(v)
adj[v].append(u)
# A utility function to print the adjacency list
# representation of graph
def printGraph(adj, V):
for v in range(V):
print("Adjacency list of vertex ",v,"\n head ")
for x in adj[v]:
print("-> ",x,end=" ")
printf()
# /* Function to return true if the graph represented
# by the adjacency list represents a ring topology
# else return false */
def checkRingTopologyUtil(adj, V, E):
# Number of edges should be equal
# to (Number of vertices - 1)
if (E != (V)):
return False
# For a graph to represent a ring topology should have
# greater than 2 nodes
if (V <= 2):
return False
vertexDegree = [0]*(V + 1)
# calculate the degree of each vertex
for i in range(V+1):
for v in adj[i]:
vertexDegree[v] += 1
# countDegree2 stores the count of
# the vertices having degree 2
countDegree2 = 0
for i in range(1, V + 1):
if (vertexDegree[i] == 2):
countDegree2 += 1
# if all three necessary conditions as discussed,
# satisfy return true
if (countDegree2 == V):
return True
else:
return False
# Function to check if the graph represents a ring topology
def checkRingTopology(adj, V, E):
isRing = checkRingTopologyUtil(adj, V, E)
if (isRing):
print("YES")
else:
print("NO" )
# Driver code
# Graph 1
V,E = 6,6
adj1 = [[] for i in range(V + 1)]
addEdge(adj1, 1, 2)
addEdge(adj1, 2, 3)
addEdge(adj1, 3, 4)
addEdge(adj1, 4, 5)
addEdge(adj1, 6, 1)
addEdge(adj1, 5, 6)
checkRingTopology(adj1, V, E)
# Graph 2
V,E = 5,4
adj2 = [[] for i in range(V + 1)]
addEdge(adj2, 1, 2)
addEdge(adj2, 1, 3)
addEdge(adj2, 3, 4)
addEdge(adj2, 4, 2)
checkRingTopology(adj2, V, E)
# This code is contributed by mohit kumar 29C#
// C# program to check if the given graph
// represents a Ring topology
using System;
using System.Collections.Generic;
class GFG
{
// A utility function to add an edge in an
// undirected graph.
static void addEdge(List> adj, int u, int v )
{
adj[u].Add(v);
adj[v].Add(u);
}
// A utility function to print the adjacency list
// representation of graph
static void printGraph(List> adj, int V)
{
for(int v = 0; v < V; ++v)
{
Console.Write("\n Adjacency list of vertex " + v + "\n head ");
foreach(int x in adj[v])
{
Console.Write(". " + x);
}
Console.WriteLine();
}
}
// Function to return true if the graph represented
// by the adjacency list represents a Ring topology
// else return false
static bool checkRingTopologyUtil(List> adj, int V, int E)
{
// Number of edges should be equal
// to Number of vertices
if (E != V)
return false;
// For a graph to represent a ring
// topology should have greater
// than 2 nodes
if (V <= 2)
return false;
int[] vertexDegree = new int[V + 1];
// Calculate the degree of each vertex
for(int i = 1; i <= V; i++)
{
foreach(int v in adj[i])
{
vertexDegree[v]++;
}
}
// countDegree2 stores the count of
// the vertices having degree 2
int countDegree2 = 0;
for(int i = 1; i <= V; i++)
{
if (vertexDegree[i] == 2)
{
countDegree2++;
}
}
// If all three necessary conditions
// as discussed, satisfy return true
if (countDegree2 == V)
{
return true;
}
else
{
return false;
}
}
// Function to check if the graph represents
// a Ring topology
static void checkRingTopology(List> adj, int V, int E)
{
bool isRing = checkRingTopologyUtil(adj, V, E);
if (isRing)
{
Console.Write("YES" + "\n");
}
else
{
Console.Write("NO" + "\n");
}
}
// Driver code
static public void Main ()
{
// Graph 1
int V = 6, E = 6;
List> adj1 = new List>();
for(int i = 0; i < V + 1; i++)
{
adj1.Add(new List() );
}
addEdge(adj1, 1, 2);
addEdge(adj1, 2, 3);
addEdge(adj1, 3, 4);
addEdge(adj1, 4, 5);
addEdge(adj1, 6, 1);
addEdge(adj1, 5, 6);
checkRingTopology(adj1, V, E);
// Graph 2
V = 6;
E = 6;
List> adj2 = new List>();
for(int i = 0; i < V + 1; i++)
{
adj2.Add(new List() );
}
addEdge(adj2, 1, 2);
addEdge(adj2, 1, 3);
addEdge(adj2, 3, 4);
addEdge(adj2, 4, 5);
checkRingTopology(adj2, V, E);
}
}
// This code is contributed by avanitrachhadiya2155 输出：
YES
NO时间复杂度：O(V + E)，其中V和E分别是图形中的顶点和边数。
如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。