utils/scrapers/prescraped/codeforces/Editorials/1942C1

Problem Credits: cry Analysis: cry
**Solution (Easy Version)**
Ignoring $$$n$$$ for now, let's just focus on the $$$x$$$ chosen vertices. Sort the $$$x$$$ vertices and connect adjacent vertices to form its own smaller polygon. By drawing out some cases or if you're familiar with triangulation (video that proves by induction), you can form $$$x - 2$$$ triangles by drawing diagonals in a polygon with $$$x$$$ vertices. If you don't know it, one possible construction that always work is fixing one vertex $$$v$$$ and drawing diagonals to every other vertex not adjacent to $$$v$$$. Now, let's consider the original $$$n$$$-sided polygon. In additional to the aforementioned construction, to close the $$$x$$$-sided polygon up: for every non-adjacent vertex in the chosen vertices based on the bigger $$$n$$$-sided polygon, draw a diagonal between them. Through this construction, we can always guarantee $$$x - 2$$$ triangles. 
However, this doesn't account for all triangles, as some triangles can form sides with the bigger $$$n$$$-sided polygon. These triangles occur exactly when two adjacent vertex in $$$x$$$ have exactly one vertex not in $$$x$$$ between them but part of the bigger polygon. This is because one side is from the diagonals we drew, and the other two sides form from the $$$n$$$-sided polygon.
Therefore, the answer is $$$x - 2$$$ + (number of adjacent chosen vertices $$$2$$$ apart). Note that the first chosen vertex is also adjacent to last chosen vertex in $$$x$$$.
**Solution (Hard Version)**
Read the solution to the easy version first.
We can now reduce the problem to the following: For every vertex we choose, we can make one more triangle. For every chosen vertices two apart, we can make one more triangle. Let's focus on the latter condition. To maximize the effect of our $$$y$$$ vertices, we want to prioritize vertices $$$v$$$ that satisfy the following: the vertex is not included in $$$x$$$ and there is a chosen vertex two apart. Let's denote such $$$v$$$ as good.
Let vertex $$$a$$$ and $$$b$$$ $$$(a < b)$$$ be adjacent chosen vertices in the $$$x$$$-sided polygon. There are $$$g = b - a - 1$$$ vertices for us to choose between them. There are two cases: if $$$g$$$ is odd, then we can choose $$$\lfloor \frac{g}{2} \rfloor$$$ vertices to form $$$\lfloor \frac{g}{2}\rfloor + 1$$$ extra triangles. This is because all of the vertices we choose will be good. if $$$g$$$ is even, then we can choose $$$\frac{g}{2}$$$ vertices to make $$$\frac{g}{2}$$$ extra triangles. This is because one of the vertices we choose will not be good. This shows that it is always optimal to process smaller and odd $$$g$$$ first. 
After processing these adjacent gaps, if we have any leftover vertices, we can just ignore them. This is because since we have maximized the number of good vertices, even though any further vertex we place will increase the answer by $$$1$$$, it will break two good vertices which will decrease the answer by $$$1$$$.