CF1685E.The Ultimate LIS Problem

It turns out that this is exactly the $100$ -th problem of mine that appears in some programming competition. So it has to be special! And what can be more special than another problem about LIS...

You are given a permutation $p_1, p_2, \ldots, p_{2n+1}$ of integers from $1$ to $2n+1$ . You will have to process $q$ updates, where the $i$ -th update consists in swapping $p_{u_i}, p_{v_i}$ .

After each update, find any cyclic shift of $p$ with $LIS \le n$ , or determine that there is no such shift. (Refer to the output section for details).

Here $LIS(a)$ denotes the length of longest strictly increasing subsequence of $a$ .

Hacks are disabled in this problem. Don't ask why.

The first line of the input contains two integers $n, q$ ( $2 \le n \le 10^5$ , $1 \le q \le 10^5$ ).

The second line of the input contains $2n+1$ integers $p_1, p_2, \ldots, p_{2n+1}$ ( $1 \le p_i \le 2n+1$ , all $p_i$ are distinct) — the elements of $p$ .

The $i$ -th of the next $q$ lines contains two integers $u_i, v_i$ ( $1 \le u_i, v_i \le 2n+1$ , $u_i \neq v_i$ ) — indicating that you have to swap elements $p_{u_i}, p_{v_i}$ in the $i$ -th update.

After each update, output any $k$ $(0 \le k \le 2n)$ , such that the length of the longest increasing subsequence of $(p_{k+1}, p_{k+2}, \ldots, p_{2n+1}, p_1, \ldots, p_k)$ doesn't exceed $n$ , or $-1$ , if there is no such $k$ .

After the first update, our permutation becomes $(5, 2, 3, 4, 1)$ . We can show that all its cyclic shifts have $LIS \ge 3$ .

After the second update, our permutation becomes $(1, 2, 3, 4, 5)$ . We can show that all its cyclic shifts have $LIS \ge 3$ .

After the third update, our permutation becomes $(1, 2, 3, 5, 4)$ . Its shift by $2$ is $(3, 5, 4, 1, 2)$ , and its $LIS = 2$ .

After the fourth update, our permutation becomes $(1, 2, 3, 4, 5)$ . We can show that all its cyclic shifts have $LIS \ge 3$ .

After the fifth update, our permutation becomes $(4, 2, 3, 1, 5)$ . Its shift by $4$ is $(5, 4, 2, 3, 1)$ , and its $LIS = 2$ .

After the fifth update, our permutation becomes $(4, 5, 3, 1, 2)$ . Its shift by $0$ is $(4, 5, 3, 1, 2)$ , and its $LIS = 2$ .