minor

Implementations/content/geometry (13)/3D/Hull3D.h

@@ -3,7 +3,11 @@
  	* are coplanar. Normals to returned faces point outwards. 
  	* If coordinates are ints at most $B$ then \texttt{T} 
  	* should be large enough to support ints on the order 
- 	* of $B^3$. Changes order of points.
+ 	* of $B^3$. Changes order of points. WARNING: If there are
+ 	* coplanar points, the number of faces that are output will
+ 	* depend on the random seed, because points that are on the
+ 	* boundary of the convex hull may or may not be included
+ 	* in the output.
  * Time: O(N^2), O(N\log N)
  * Source: 
  	* KACTL
@@ -63,10 +67,9 @@ V<F> hull3d(vP3& p) {
 }
 V<F> hull3dFast(vP3& p) {
 	prep(p); int N = sz(p); V<F> hull; 
-	vb active; V<vi> rvis; V<AR<pi,3>> other;
-	// whether face is active
-	// points visible from each face
-	// other face adjacent to each edge of face
+	vb active; // whether face is active
+	V<vi> rvis; // points visible from each face
+	V<AR<pi,3>> other; // other face adjacent to each edge of face
 	V<vi> vis(N); // faces visible from each point
 	auto ad = [&](int a, int b, int c) { 
 		hull.pb({a,b,c}); active.pb(1); rvis.eb(); other.eb(); };

Implementations/content/graphs (12)/Matching/UnweightedMatch.h

@@ -15,7 +15,7 @@
 
 template<int SZ> struct UnweightedMatch {
 	int match[SZ], N; vi adj[SZ];
-	void ae(int u, int v) { adj[u].pb(v), adj[v].pb(u); }
+	void ae(int u, int v) { assert(u != v); adj[u].pb(v), adj[v].pb(u); }
 	queue<int> q;
 	int par[SZ], vis[SZ], orig[SZ], aux[SZ];
 	void augment(int u, int v) { // toggle edges on u-v path

Other Implementations/06 - DP/LIS (6).cpp

@@ -4,10 +4,12 @@
  * Verification: https://cses.fi/problemset/task/1145/
  */
 
-vi bes = {INT_MIN}; // min last term of increasing sequence with i terms
-
-void add(int x) { // add terms of sequence one by one
-    int lo = lb(all(bes),x)-begin(bes);
-    if (lo == sz(bes)) bes.pb(0);
-    bes[lo] = x; // sz(bes)-1 is your current answer
-}
+int lis(vi v) {
+	vi min_last{INT_MIN}; // min last term of increasing sequence with i terms
+	for (int x: v) {
+		int lo = lb(all(min_last),x)-begin(min_last);
+		if (lo == sz(min_last)) min_last.pb(0);
+		min_last[lo] = x;
+	}
+	return sz(min_last)-1;
+}