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; }