Complexity of Hopcroft-Karp I have a rather basic question about the number of operations taken by the Hopcroft-Karp algorithm for finding a maximum matching in a bipartite graph. It is commonly reported as $O(m \sqrt{n})$ where $m$ is the number of edges in the graph and $n$ is the number of vertices(for example, see wikipedia or this book, among many examples - I cannot post more than two links due to reputation constraints). 

What bothers me about this is that if $m$ is small (say $m=n^{1/4}$) this seems to require less than $n$ iterations, which is odd: surely you have to at least look at every node, i.e., just to see if it has any incident edges or not?

Having looked at Hopcroft and Karp's original paper, I see they quote a running time of $O((m+n)\sqrt{n})$ which makes more sense. 

So my question is: what is the reason why the running time of Hopcroft-Karp is usually quoted as $O(m \sqrt{n})$? Does this reflect an assumption that the graph is connected (I never see this stated explicitly)?
The correct running time is indeed probably $O((m+n)\sqrt{n})$. However, this is a mouthful, and the expression $O(m\sqrt{n})$ looks nicer and is also more succinct. In most cases, $m \geq n/2$, since otherwise there is an isolated vertex. In fact, by finding isolated vertices, we can improve the running time of the algorithm to $O(n + m\sqrt{n})$.

An alternative way of stating the running time would be $O(m\sqrt{n})$, assuming $m \geq n$. The case of small $m$ is not so interesting (since then there are isolated vertices which can be eliminated), and so it is usually assumed that $m = \Omega(n)$.