revlog: optimize ancestors() to not check filtered revisions for each

While reviewing the Rust implementation, I noticed iter(ancestors) doesn't
need to check filtering state for each parent revision. And doing that appears
to have some measurable perf win.

  $ hg perfancestors -R mercurial
  (orig) wall 0.038093 comb 0.040000 user 0.040000 sys 0.000000 (best of 100)
  (this) wall 0.024795 comb 0.020000 user 0.020000 sys 0.000000 (best of 117)

mercurial/revlog.py

@@ -648,6 +648,9 b' class revlog(object):'
 
         return entry[5], entry[6]
 
+    # fast parentrevs(rev) where rev isn't filtered
+    _uncheckedparentrevs = parentrevs
+
     def node(self, rev):
         try:
             return self.index[rev][7]
@@ -747,8 +750,14 b' class revlog(object):'
 
         See the documentation for ancestor.lazyancestors for more details."""
 
-        return ancestor.lazyancestors(self.parentrevs, revs, stoprev=stoprev,
-                                      inclusive=inclusive)
+        # first, make sure start revisions aren't filtered
+        revs = list(revs)
+        checkrev = self.node
+        for r in revs:
+            checkrev(r)
+        # and we're sure ancestors aren't filtered as well
+        return ancestor.lazyancestors(self._uncheckedparentrevs, revs,
+                                      stoprev=stoprev, inclusive=inclusive)
 
     def descendants(self, revs):
         return dagop.descendantrevs(revs, self.revs, self.parentrevs)