I have to implement BFS algorithm for connected and disconnected graphs that returns the order of visited vertices. For connected graph everytihing is ok, but for disconnected one I have to do it in such a way that every vertex is visited also those in other connected components. In each component it should be done in according to BFS. First component start from the vertex start. I tried few things but it didn't work. If somebody could give me some hints, I would be grateful. I have also methods given below.

int areConnected(int,int)
Returns:
1 if given vertices are connected by an egde
0 if the vertices are not connected
-1 if the vertices do not exist

int getVerticesCount()
Returns number of vertices

int EdgesCount()
Returns number of edges

int getDegree(int)
Returns:
degree of the given vertex
-1 if no such vertex exists.

list<int>* getNeighbors(int)
Returns:
a pointer to the list of neighbors of a given vertex (for isolated vertex list is empty)
null if no such vertex exists

Code:
list<int>  BFS(int start)
{
        list<int> order; // order of vertices
        bool *visited=new bool[getVerticesCount()];
        queue<int> kolejka;
        for(int i=0;i<getVerticesCount();++i)
        visited[i]=false;
        visited[start]=true;
        kolejka.push(start);
        while(!kolejka.empty())
        {            
            int vertex=kolejka.front();
            kolejka.pop();
            order.push_back(vertex);
            for(int i=0;i<getVerticesCount();++i)
            {    
                if(i==vertex)
                    continue;
                if(areConnected(i,vertex)==1)
                {
                    if(visited[i]==false)
                    {
                        visited[i]=true;
                        kolejka.push(i);
                    }
                }
                else 
                    continue;
            }
        }
        delete [] visited;
        visited=NULL;
        return order;
}