使用 BFS
检测有向图中的周期
原文： https://www.geeksforgeeks.org/detect-cycle-in-a-directed-graph-using-bfs/
给定有向图，请检查该图是否包含循环。 如果给定图包含至少一个循环，则函数应返回 true，否则返回 false。 例如，下图包含两个循环 0-> 1-> 2-> 3-> 0 和 2-> 4-> 2，因此您的函数必须返回 true。
我们讨论了一种基于 DFS 的解决方案，以检测有向图中的周期。 在这篇文章中，讨论了基于 BFS 的解决方案。
的想法是简单地将 Kahn 的算法用于拓扑排序
使用 BFS 在有向图中检测循环的步骤。
步骤 1：计算图中存在的每个顶点的入度（传入边数），并将访问的节点数初始化为 0。
步骤- 2：选取度数为 0 的所有顶点并将它们添加到队列中（入队操作）
步骤 3：从队列中移除一个顶点（出队操作），然后 。
-
访问的节点数增加 1。
-
将其所有相邻节点的度数减少 1。
-
如果将相邻节点的度降低到零，则将其添加到队列中。
步骤 4：重复步骤 3，直到队列为空。
步骤 5：如果访问的节点数是而不是等于图中有循环的节点数，否则返回。
如何找到每个节点的度数？
有两种计算每个顶点的度数的方法：
取一个度数数组，该数组将跟踪
- 遍历边和 只需将目标节点的计数器加 1。
for each node in Nodes
indegree[node] = 0;
for each edge(src,dest) in Edges
indegree[dest]++
时间复杂度：O(V + E)
- 遍历每个节点的列表，然后将与其连接的所有节点的入度增加 1。
for each node in Nodes
If (list[node].size()!=0) then
for each dest in list
indegree[dest]++;
时间复杂度：外层 for 循环将执行 V 次，内层 for 循环将执行 E 次，因此总时间复杂度为O(V + E)。
该算法的总时间复杂度为O(V + E)
C++
// A C++ program to check if there is a cycle in
// directed graph using BFS.
#include <bits/stdc++.h>
using namespace std;
// Class to represent a graph
class Graph {
int V; // No. of vertices'
// Pointer to an array containing adjacency lisr
list<int>* adj;
public:
Graph(int V); // Constructor
// function to add an edge to graph
void addEdge(int u, int v);
// Returns true if there is a cycle in the graph
// else false.
bool isCycle();
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int u, int v)
{
adj[u].push_back(v);
}
// This function returns true if there is a cycle
// in directed graph, else returns false.
bool Graph::isCycle()
{
// Create a vector to store indegrees of all
// vertices. Initialize all indegrees as 0.
vector<int> in_degree(V, 0);
// Traverse adjacency lists to fill indegrees of
// vertices. This step takes O(V+E) time
for (int u = 0; u < V; u++) {
for (auto v : adj[u])
in_degree[v]++;
}
// Create an queue and enqueue all vertices with
// indegree 0
queue<int> q;
for (int i = 0; i < V; i++)
if (in_degree[i] == 0)
q.push(i);
// Initialize count of visited vertices
int cnt = 0;
// Create a vector to store result (A topological
// ordering of the vertices)
vector<int> top_order;
// One by one dequeue vertices from queue and enqueue
// adjacents if indegree of adjacent becomes 0
while (!q.empty()) {
// Extract front of queue (or perform dequeue)
// and add it to topological order
int u = q.front();
q.pop();
top_order.push_back(u);
// Iterate through all its neighbouring nodes
// of dequeued node u and decrease their in-degree
// by 1
list<int>::iterator itr;
for (itr = adj[u].begin(); itr != adj[u].end(); itr++)
// If in-degree becomes zero, add it to queue
if (--in_degree[*itr] == 0)
q.push(*itr);
cnt++;
}
// Check if there was a cycle
if (cnt != V)
return true;
else
return false;
}
// Driver program to test above functions
int main()
{
// Create a graph given in the above diagram
Graph g(6);
g.addEdge(0, 1);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(3, 4);
g.addEdge(4, 5);
if (g.isCycle())
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program to check if there is a cycle in
// directed graph using BFS.
import java.io.*;
import java.util.*;
class GFG
{
// Class to represent a graph
static class Graph
{
int V; // No. of vertices'
// Pointer to an array containing adjacency list
Vector<Integer>[] adj;
@SuppressWarnings("unchecked")
Graph(int V)
{
// Constructor
this.V = V;
this.adj = new Vector[V];
for (int i = 0; i < V; i++)
adj[i] = new Vector<>();
}
// function to add an edge to graph
void addEdge(int u, int v)
{
adj[u].add(v);
}
// Returns true if there is a cycle in the graph
// else false.
// This function returns true if there is a cycle
// in directed graph, else returns false.
boolean isCycle()
{
// Create a vector to store indegrees of all
// vertices. Initialize all indegrees as 0.
int[] in_degree = new int[this.V];
Arrays.fill(in_degree, 0);
// Traverse adjacency lists to fill indegrees of
// vertices. This step takes O(V+E) time
for (int u = 0; u < V; u++)
{
for (int v : adj[u])
in_degree[v]++;
}
// Create an queue and enqueue all vertices with
// indegree 0
Queue<Integer> q = new LinkedList<Integer>();
for (int i = 0; i < V; i++)
if (in_degree[i] == 0)
q.add(i);
// Initialize count of visited vertices
int cnt = 0;
// Create a vector to store result (A topological
// ordering of the vertices)
Vector<Integer> top_order = new Vector<>();
// One by one dequeue vertices from queue and enqueue
// adjacents if indegree of adjacent becomes 0
while (!q.isEmpty())
{
// Extract front of queue (or perform dequeue)
// and add it to topological order
int u = q.poll();
top_order.add(u);
// Iterate through all its neighbouring nodes
// of dequeued node u and decrease their in-degree
// by 1
for (int itr : adj[u])
if (--in_degree[itr] == 0)
q.add(itr);
cnt++;
}
// Check if there was a cycle
if (cnt != this.V)
return true;
else
return false;
}
}
// Driver Code
public static void main(String[] args)
{
// Create a graph given in the above diagram
Graph g = new Graph(6);
g.addEdge(0, 1);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(3, 4);
g.addEdge(4, 5);
if (g.isCycle())
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by
// sanjeev2552
Python
# A Python3 program to check if there is a cycle in
# directed graph using BFS.
import math
import sys
from collections import defaultdict
# Class to represent a graph
class Graph:
def __init__(self,vertices):
self.graph=defaultdict(list)
self.V=vertices # No. of vertices'
# function to add an edge to graph
def addEdge(self,u,v):
self.graph[u].append(v)
# This function returns true if there is a cycle
# in directed graph, else returns false.
def isCycleExist(n,graph):
# Create a vector to store indegrees of all
# vertices. Initialize all indegrees as 0\.
in_degree=[0]*n
# Traverse adjacency lists to fill indegrees of
# vertices. This step takes O(V+E) time
for i in range(n):
for j in graph[i]:
in_degree[j]+=1
# Create an queue and enqueue all vertices with
# indegree 0
queue=[]
for i in range(len(in_degree)):
if in_degree[i]==0:
queue.append(i)
# Initialize count of visited vertices
cnt=0
# One by one dequeue vertices from queue and enqueue
# adjacents if indegree of adjacent becomes 0
while(queue):
# Extract front of queue (or perform dequeue)
# and add it to topological order
nu=queue.pop(0)
# Iterate through all its neighbouring nodes
# of dequeued node u and decrease their in-degree
# by 1
for v in graph[nu]:
in_degree[v]-=1
# If in-degree becomes zero, add it to queue
if in_degree[v]==0:
queue.append(v)
cnt+=1
# Check if there was a cycle
if cnt==n:
return False
else:
return True
# Driver program to test above functions
if __name__=='__main__':
# Create a graph given in the above diagram
g=Graph(6)
g.addEdge(0,1)
g.addEdge(1,2)
g.addEdge(2,0)
g.addEdge(3,4)
g.addEdge(4,5)
if isCycleExist(g.V,g.graph):
print("Yes")
else:
print("No")
# This Code is Contributed by Vikash Kumar 37
C
// C# program to check if there is a cycle in
// directed graph using BFS.
using System;
using System.Collections.Generic;
class GFG{
// Class to represent a graph
public class Graph
{
// No. of vertices'
public int V;
// Pointer to an array containing
// adjacency list
public List<int>[] adj;
public Graph(int V)
{
// Constructor
this.V = V;
this.adj = new List<int>[V];
for (int i = 0; i < V; i++)
adj[i] = new List<int>();
}
// Function to add an edge to graph
public void addEdge(int u, int v)
{
adj[u].Add(v);
}
// Returns true if there is a cycle in the
// graph else false.
// This function returns true if there is
// a cycle in directed graph, else returns
// false.
public bool isCycle()
{
// Create a vector to store indegrees of all
// vertices. Initialize all indegrees as 0.
int[] in_degree = new int[this.V];
// Traverse adjacency lists to fill indegrees
// of vertices. This step takes O(V+E) time
for(int u = 0; u < V; u++)
{
foreach(int v in adj[u])
in_degree[v]++;
}
// Create an queue and enqueue all
// vertices with indegree 0
Queue<int> q = new Queue<int>();
for(int i = 0; i < V; i++)
if (in_degree[i] == 0)
q.Enqueue(i);
// Initialize count of visited vertices
int cnt = 0;
// Create a vector to store result
// (A topological ordering of the
// vertices)
List<int> top_order = new List<int>();
// One by one dequeue vertices from
// queue and enqueue adjacents if
// indegree of adjacent becomes 0
while (q.Count != 0)
{
// Extract front of queue (or perform
// dequeue) and add it to topological
// order
int u = q.Peek();
q.Dequeue();
top_order.Add(u);
// Iterate through all its neighbouring
// nodes of dequeued node u and decrease
// their in-degree by 1
foreach(int itr in adj[u])
if (--in_degree[itr] == 0)
q.Enqueue(itr);
cnt++;
}
// Check if there was a cycle
if (cnt != this.V)
return true;
else
return false;
}
}
// Driver Code
public static void Main(String[] args)
{
// Create a graph given in the above diagram
Graph g = new Graph(6);
g.addEdge(0, 1);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(3, 4);
g.addEdge(4, 5);
if (g.isCycle())
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Princi Singh
Output:
Yes
时间复杂度：O(V + E)
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