If you take n as packages (vertices) you can have n² dependencies (edges), each of which needs to be looked at once no matter what. So if you count that way n²

If you take n as packages (vertices) you can have n² dependencies (edges), each of which needs to be looked at once no matter what. So if you count that way n² is unavoidable in any dependency sorting. (Usually for such algorithms I see n = |v| + |e|, or max(|v|, |e|), or just left as |v| and |e|.)