I am facing a small doubt in SPFA problem. My algorithm not working for some test cases

Code:

bool SPFA(int source) //Returns true if there is a negative cycle reachable from source
{
	queue<int> q;


	for(int i=0;i<=n;i++)
	{
		cnt[i]=0;
		dist[i]=INF;
		inqueue[i] = false;
		
	}

	dist[source]=0;
	q.push(source);
	inqueue[source]=true;

	while(!q.empty())
	{
		int v=q.front();
		q.pop();
		inqueue[v]=false;

		for(auto edge:g[v])
		{
			int to=edge.first;
			int w=edge.second;

			if(dist[v] + w < dist[to])
			{
				dist[to] = dist[v] + w;
				//dist[to] = max(dist[to], -INF);
				if(!inqueue[to])
				{
					q.push(to);
					inqueue[to]=true;
					cnt[to]++;
					if(cnt[to]>n){
						

						   return true;
					}
				}
			}
		}
	}
return false;

	}



Here in my code for a test case  :  N = 4 src = 1 
                              u  v   w
            
                               19 20 -36
                               20 19 34
                                 

Negative cycle has to be 19->20->19

As queue starts from src(1)

---> No edge from src so it can't visit adjacent vertices so we can't relax edges connected to node.

Can we add an imaginary edge in the graph so that we can traverse ?

Any help would be encouraging.

After constructing bridge tree, we have bi-connected components as nodes and bridges as edges. Is there any possible way to print every bicomponent which is a node in bridge tree.

eg:

Its bridge tree is

Can we print all biconnected components from bridge tree itself?

Here in the above graph : Comp1 -> (1-2-3-4) Comp2--> (5-6-7) Comp3-->(5-9) Comp4---> (7-8)