Go4Expert - Breath-First Search algorithm C++

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;

}