Reputation: 137
I have question similar - same - as this one. So I want to know how to not only detect cycle but also print out vertices which this cycle contains. I tried ways which were mentioned in the question above, but I must have done something wrong, why it doesn't work for my. Also my program checks just if one specific vertex makes cycle. My code is here:
#include<iostream>
#include <list>
using namespace std;
class Graph
{
int V;
list<int> *adj;
public:
Graph(int V);
void addEdge(int v, int w);
bool Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent, int index);
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v);
}
bool Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent, int index)
{
visited[v] = true;
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
{
if (!visited[*i])
{
if (isCyclicUtil(*i, visited, cycleVertices, v, index)) {
if (index <= 1 || cycleVertices[0] != cycleVertices[index - 1])
cycleVertices[index++] = *i;
return true;
}
}
else if (*i != parent) {
cycleVertices[index++] = *i;
return true;
}
}
return false;
}
int main()
{
bool *visited = new bool[5];
for (int i = 0; i < 5; i++)
visited[i] = false;
int cycleVertices[5];
for (int i = 0; i < 5; i++)
cycleVertices[i] = -1;
Graph g1(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.isCyclicUtil(4, visited, cycleVertices, -1, 0) ? cout << "Graph contains cycle\n" :
cout << "Graph doesn't contain cycle\n";
int x = 0;
while (cycleVertices[x] != -1)
cout << cycleVertices[x++] << " ";
return 0;
}
Upvotes: 1
Views: 5662
Reputation: 379
Here is my DFS solution in Python
#program to print all nodes included in the cycle in the given undirected graph
import collections
edges = [[1, 2], [2, 3], [1, 3], [2, 4], [4, 5], [5, 6], [4, 6]]
n = 6
parent = [0] * (n + 1)
color = [0] * (n + 1)
mark = [0] * (n + 1)
cycleno = 0
graph = collections.defaultdict(set)
for i, j in edges:
graph[i].add(j)
graph[j].add(i)
def dfs(u, v):
global cycleno
if color[u] == 2: #node explore complete
return
elif color[u] == 1: #cycle found
cycleno += 1
cur = v
mark[cur] = cycleno
while cur != u:
cur = parent[cur]
mark[cur] = cycleno
else:
parent[u] = v
color[u] = 1
for nei in graph[u]:
if nei == parent[u]: continue
dfs(nei, u)
color[u] = 2 #exploration for this node completed
dfs(1, 0)
print(mark)
Output:
[0, 1, 1, 1, 2, 2, 2]
which means nodes 1, 2, 3 and 4, 5, 6 belongs to the cycles respectively.
Reference: https://www.tutorialspoint.com/print-all-the-cycles-in-an-undirected-graph-in-cplusplus
Upvotes: 0
Reputation: 137
I've found a solution. I've tried j_random_hacker's solution in this post and it didn't work. But problem was with indexing in cycleVertices in my code. Variable index was always same. So I've added a new attribute index in the class Graph and now it works. So here is the edited code:
#include<iostream>
#include <list>
#define FINISHED -1
#define NOCYCLE -2
using namespace std;
class Graph
{
int V;
int index;
list<int> *adj;
public:
Graph(int V);
void addEdge(int v, int w);
void set_index();
int Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent);
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
this->index = 0;
}
void Graph::set_index()
{
this->index = 0;
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v);
}
int Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent)
{
visited[v] = true;
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
{
if (!visited[*i])
{
int result = isCyclicUtil(*i, visited, cycleVertices, v);
if (result == FINISHED)
return FINISHED;
else if (result != NOCYCLE) {
cycleVertices[index++] = v;
if (result == v)
return FINISHED;
else
return result;
}
}
else if (*i != parent) {
return *i;
}
}
return NOCYCLE;
}
int main()
{
bool *visited = new bool[4];
for (int i = 0; i < 4; i++)
visited[i] = false;
int cycleVertices[4];
for (int i = 0; i < 4; i++)
cycleVertices[i] = -1;
Graph g1(4);
g1.addEdge(0, 1);
g1.addEdge(1, 2);
g1.addEdge(2, 3);
g1.addEdge(3, 0);
g1.isCyclicUtil(3, visited, cycleVertices, -1) ? cout << "Graph contains cycle\n" :
cout << "Graph doesn't contain cycle\n";
int x = 0;
while (cycleVertices[x] != -1)
cout << cycleVertices[x++] << " ";
return 0;
}
Upvotes: 1