CareerCup

int n;
vector<int> g[MAXN];
bool used[MAXN];
vector<int> comp;
int res = 0; //Number Of Connected Components

void dfs (int v) {
used[v] = true;
comp.push_back (v);
for (size_t i=0; i<g[v].size(); ++i) {
int to = g[v][i];
if (! used[to])
dfs (to);
}
}

void find_comps() {
for (int i=0; i<n; ++i)
used[i] = false;
for (int i=0; i<n; ++i)
if (! used[i]) {
comp.clear();
dfs (i);

res++;
}
}

This can be done using Disjoint Set.
1. For each vertex v, call MakeSet(v)
2. For each edge vertices, u & v, UnionSet(u, v). Modify UnionSet to return parent, if parent is a new one, increment count
3. Count has number of connected components

//Simple DFS approach
int components(){
	int count = 0;
	for(Node n : Graph.vertices()){
		if(!visited(n)){
			count++;
			DFS(n);
		}
	}
	return count;
}

void DFS(Node node){
	visit(node);
	for(Node adj : node.getAdjacentNodes()){
		DFS(adj);
	}
}

//To implement visited we can use an array or Map\Dictionary