Disclaimer: A lot of this post was copied from my answer to a similar sort of game.
Summary
It turns out that this game is very closely related to the Octal Game with code "0.007". Using the Sprague-Grundy Theorem combined with a winning strategy for Nim, it is relatively straightforward to analyze small games like this (e.g. your game for small $n$), and there are theorems which give efficient algorithms to learn about this in the medium range. Some Octal games can be efficiently solved even in large cases, but this particular game is not one of them.
It turns out that $B$ has a winning strategy for $N={0, 1, 2, 8, 14, 24, 32, 34, 46, 56, 66, 78, 88, 100, 112, 120, 132, 134, 164, 172, 186, 196, 204, 284, 292, 304, 358, 1048, 2504, 2754, 2914, 3054, 3078, 7252, 7358, 7868, 16170}$ and $A$ has a winning strategy in all other cases up to $2^{28}-1$. But I believe the general result is still open.
Connecting this game to existing theory
First note that you shouldn't be able to draw a curve through points as then drawing a curve through all of the points would be a winning move all of the time. To connect this game to existing general theory, we can reconceptualize it in the way that Lærne did: After you draw a curve, you remove three points from a component, and split the remainder into two components (where one or both of those may be empty). That is traditionally encoded in a "number" with Octal digits. In this case, the code is $0.007$, where the $0$s indicate that you can't remove only one or two points, and the three bits of the $7$ indicate that you can leave zero or one or two components when you remove three points from a component.
Who wins?
Since the original question was "who has a winning strategy?", rather than "what is the winning strategy, I won't re-present an introduction to the relevant theorems in detail.
If you don't already know about the strategy for Nim or the Sprague-Grundy Theorem and how they can be applied to similar games, you can read one or more of the following:
Very briefly, positions in games like this act in combinations like single heaps of Nim, where the size of the corresponding Nim heap is the least number that's not a size of a Nim heap corresponding to a position you can move to. The strategy for Nim tells you that combining games with known equivalent Nim boils down to bitwise XOR (aka "adding in base $2$ without carrying").
Because of the strategy for Nim (how Grundy values combine), and a position in this game is a combination of separate components, it suffices to know the Grundy/Nim values for a single "heap" (in our context, a single component of points). There are tricks to make this calculation more efficient than the naive method, some of which are covered in the graduate textbook "Combinatorial Game Theory" by Aaron N. Siegel. For some similar games, the Nim values are known to be eventually periodic (by a theorem that says they will be if they look periodic long enough), but according to Flammenkamp, for $0.007$, $2^{28}$ values have been calculated with no periodic behavior, although the values $N$ for which $B$ wins appear to stop at $16170$.
Aaron Siegel's program Combinatorial Game Suite allows efficient calculation of middling values (but probably wouldn't get up to $2^{28}$. For example, the lines hr:=game.heap.HeapRules.MakeRules("0.007")
and hr.NimValues(100000)
takes about a minute and a half on my machine to produce the list of Nim values for $N=0$ to $N=99999$.
For example, the Nim values from $N=0$ to $N=499$ are as follows (remember that $N=0$ corresponds to a win for $B$) $0,0,0,1,1,1,2,2,0,3,3,1,1,1,0,4,3,3,3,2,2,2,4,4,0,5,5,2,2,2,3,3,0,5,0,1,1,1,3,3,3,5,6,4,4,1,0,5,5,6,6,2,7,7,7,8,0,1,9,2,7,2,3,3,3,9,0,5,4,4,8,6,6,2,7,1,1,1,0,5,5,9,3,1,8,2,8,5,0,1,1,12,2,2,7,3,3,9,4,4,0,11,3,3,3,9,2,2,8,1,3,5,0,9,12,2,6,13,13,5,0,1,1,4,11,7,7,10,3,4,1,4,0,5,0,3,3,6,7,2,14,13,10,4,12,9,2,2,3,3,6,9,9,1,16,4,8,3,3,2,15,1,1,4,0,5,5,16,6,6,6,8,0,16,5,4,4,17,2,2,7,14,6,10,12,1,0,16,13,3,6,2,7,7,8,1,0,5,17,2,12,15,3,11,0,19,18,12,4,16,17,2,2,21,6,9,4,19,5,5,17,10,3,6,19,2,7,8,4,1,9,12,7,2,13,6,3,19,5,9,4,8,8,17,17,2,15,18,1,1,8,5,21,16,21,3,19,19,13,5,18,1,4,17,7,2,7,6,3,19,12,5,5,16,16,6,17,19,7,7,18,1,4,17,0,9,16,3,3,14,13,22,0,1,15,24,17,2,6,18,3,4,19,19,0,8,21,16,3,15,7,26,18,13,1,1,17,9,2,21,2,6,22,19,9,5,16,4,16,20,3,7,18,23,22,8,20,5,16,21,15,6,10,19,18,18,18,4,4,17,17,7,2,3,23,19,9,5,0,16,16,3,17,30,2,18,18,8,4,17,17,9,27,6,10,19,19,14,9,9,4,20,17,14,11,7,18,6,19,19,5,13,16,16,10,6,19,19,23,18,4,4,17,12,12,14,10,6,3,19,5,9,5,21,16,20,6,7,7,18,30,13,13,17,12,21,15,10,3,19,22,18,8,4,32,17,17,11,14,6,26,24,12,5,9,16,16,6,7,7,7,18,18,8,4,17,20,7,16,10,10,22,19,22,9,23,4,13,17,20,7,11,23,23,4,5,9,5,16,16,10,17,10,22,18,23,8,4,17,17,20,16,32,13,19,19,33,5,5,24$
Best Answer
First note that we can think of the row as a row of $0$'s and $1$'s because we only care about pairity.
Let $n$ be the length of the row.
Claim:
$1)$. if $n$ is even and $A$ begins, than $A$ wins.
$2)$. if $n$ is odd and $B$ begins, than $A$ wins.
Note that $1)$ is enough for us for the problem, but we will prove both $1$ and $2$ simultaneously by induction on $n$.
The base cases $n=1,2,3$ are trivial. Suppose now the claim holds for $1,2,...,n-1$ and let's prove it for $n$.
If $n$ is even, the row is of the form $1,0,1,0,1,0,....,1,0$ and we may take the last pair of $(1,0)$ and make it a $1$. By induction hypothesis, since now it is $B$'s turn, $A$ wins.
If $n$ is odd and $B$ begins, the row is of the form $1,0,1,0,1,0,....,1,0,1$.
Player $B$ might pick a pair of the form $(1,0)$ or a pair of the form $(0,1)$, and replace this pair with a $0$ or a $1$. Note that doesn't matter what pair he chose to replace, there is either a $1$ right before it or a $1$ right after it: there is a $1$ after each pair of the form $(1,0)$ and there is a $1$ before each pair of the form $(0,1)$. Therefore the two numbers $B$ chose to replace are part of $3$ consecutive numbers in the row, which are $1,0,1$ (and $B$ chose either the first two or the last two of those three). Doesn't matter what $B$ did, we can make the $1,0,1$ that $B$ chose his pair from a $1$. For example, if $B$ chose the first two and made them a $0$, the $1,0,1$ became a $0,1$, which $A$ can make a $1$ from. Therefore, after $A$'s second turn, the difference from the original row is that a $1,0,1$ was replaced by a $1$. Now, since it is $B$'s turn now, by induction hypothesis $A$ wins.