Pokémon Army (hard version)
Contest: Codeforces Round 672 (Div. 2)
Difficulty rating: 2100
Tags: data structures, divide and conquer, dp, greedy, implementation

This is the hard version of the problem. The difference between the versions is that the easy version has no swap operations. You can make hacks only if all versions of the problem are solved.

Pikachu is a cute and friendly pokémon living in the wild pikachu herd.

But it has become known recently that infamous team R wanted to steal all these pokémon! Pokémon trainer Andrew decided to help Pikachu to build a pokémon army to resist.

First, Andrew counted all the pokémon — there were exactly $$$n$$$ pikachu. The strength of the $$$i$$$-th pokémon is equal to $$$a_i$$$, and all these numbers are distinct.

As an army, Andrew can choose any non-empty subsequence of pokemons. In other words, Andrew chooses some array $$$b$$$ from $$$k$$$ indices such that $$$1 \le b_1 < b_2 < \dots < b_k \le n$$$, and his army will consist of pokémons with forces $$$a_{b_1}, a_{b_2}, \dots, a_{b_k}$$$.

The strength of the army is equal to the alternating sum of elements of the subsequence; that is, $$$a_{b_1} - a_{b_2} + a_{b_3} - a_{b_4} + \dots$$$.

Andrew is experimenting with pokémon order. He performs $$$q$$$ operations. In $$$i$$$-th operation Andrew swaps $$$l_i$$$-th and $$$r_i$$$-th pokémon.

Andrew wants to know the maximal stregth of the army he can achieve with the initial pokémon placement. He also needs to know the maximal strength after each operation.

Help Andrew and the pokémon, or team R will realize their tricky plan!

Input format:
Each test contains multiple test cases.

The first line contains one positive integer $$$t$$$ ($$$1 \le t \le 10^3$$$) denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 3 \cdot 10^5, 0 \le q \le 3 \cdot 10^5$$$) denoting the number of pokémon and number of operations respectively.

The second line contains $$$n$$$ distinct positive integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le n$$$) denoting the strengths of the pokémon.

$$$i$$$-th of the last $$$q$$$ lines contains two positive integers $$$l_i$$$ and $$$r_i$$$ ($$$1 \le l_i \le r_i \le n$$$) denoting the indices of pokémon that were swapped in the $$$i$$$-th operation.

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$, and the sum of $$$q$$$ over all test cases does not exceed $$$3 \cdot 10^5$$$.

Output format:
For each test case, print $$$q+1$$$ integers: the maximal strength of army before the swaps and after each swap.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
Let's look at the third test case:

Initially we can build an army in such way: [1 2 5 4 3 6 7], its strength will be $$$5-3+7=9$$$.

After first operation we can build an army in such way: [2 1 5 4 3 6 7], its strength will be $$$2-1+5-3+7=10$$$.

After second operation we can build an army in such way: [2 1 5 4 3 7 6], its strength will be $$$2-1+5-3+7=10$$$.

After third operation we can build an army in such way: [2 1 4 5 3 7 6], its strength will be $$$2-1+5-3+7=10$$$.

After forth operation we can build an army in such way: [1 2 4 5 3 7 6], its strength will be $$$5-3+7=9$$$.

After all operations we can build an army in such way: [1 4 2 5 3 7 6], its strength will be $$$4-2+5-3+7=11$$$.

Example input:
3
3 1
1 3 2
1 2
2 2
1 2
1 2
1 2
7 5
1 2 5 4 3 6 7
1 2
6 7
3 4
1 2
2 3

Example output:
3
4
2
2
2
9
10
10
10
9
11

Editorial:
Let's give a solution for a fixed array and then prove its optimality.
Let us name the element $$$a_i$$$ a local maximum if $$$a_i > a_{i-1}$$$ and $$$a_i > a_{i+1}$$$. Similarly let's call the element $$$a_i$$$ a local minimum if $$$a_i < a_{i-1}$$$ and $$$a_i < a_{i+1}$$$. If any of the $$$a_{i-1}$$$ or $$$a_{i+1}$$$ does not exist in the definitions above, we assume it is equal to $$$-\infty$$$.
Note that the optimal subsequence will always be an odd length (otherwise we can delete the last element and increase the response). Elements with odd numbers shall be located at local maximums, and elements with even numbers — at local minimums.
It is not difficult to see that the first local maximum is always placed earlier than the first local minimum (otherwise it would happen that the initial permutation decreases from the first element to the local minimum, in which case the first element itself is the local maximum). Similarly, you may notice that the last local maximum always costs later than the last local minimum. Given that the local maximums and minimums alternate, you can simply take a subsequence of all the local maximums and minimums and get the best response.
Let's show that this construction is alpways optimal. Let's start with the case when an element with an odd number is not a local maximum. In this case, it shall be replaced with a bigger neighbor, and if the bigger neighbor is already in the sub-set, just delete both of these elements. After that, the response shall always increase. The same shall apply if the even-numbered element is not a local minimum. In this case, it may still happen that we cannot move the element downwards because it is on the edge. But then it is the last one in the subsequence, and it can be easily removed.
Thus, it is optimal to take only local highs and lows into the subsequence (considering that highs are on odd positions and lows — on even positions). It remains to be shown that it is profitable to take all local maximums and minimums. Indeed, if not all of them are involved, then there is a pair of standing local highs and lows. By adding them to the subsequence, we will increase the answer.
Okay. We know how to solve a problem for an initial array by reducing it to the sum of all local maximums and minimums. We will now learn how to process requests quickly. To do this, we will store whether an element is a local minimum or maximum and recalculate this information when exchanging elements. Suddenly it turns out that a single request will change the state of no more than six elements, so we can easily recalculate the response for $$$O(1)$$$ per request.
Thus, we have a solution with an asymptotic $$$O(n + q)$$$ in time.