Pokémon Army (hard version) (Codeforces 1420C2)
Tags: Greedy, divide and conquer

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

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 $$$4-2+5-3+7=11$$$.

Sample Inputs:
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

Sample Outputs:
3
4
2
2
2
9
10
10
10
9
11

Solution (Python):
import sys

def solve():
    z = sys.stdin.readline
    o = []
    y = lambda a, b: a - b if a - b > 0 else 0
    for _ in range(int(z())):
        (n, q) = map(int, z().split())
        p = [0] + [*map(int, z().split())]
        s = m = 0
        for i in p:
            if i > m:
                s += i - m
            m = i
        o.append(s)
        for i in range(q):
            (a, b) = map(int, z().split())
            if a == b:
                o.append(s)
                continue
            (va, vb) = (p[a], p[b])
            va1 = p[a - 1]
            c = b < n
            if c:
                vb1 = p[b + 1]
            (p[a], p[b]) = (vb, va)
            lv = nv = 0
            if b - a < 2:
                if vb > va:
                    lv += vb - va
                if va > va1:
                    lv += va - va1
                if c and vb1 > vb:
                    lv += vb1 - vb
                if va > vb:
                    nv += va - vb
                if vb > va1:
                    nv += vb - va1
                if c and vb1 > va:
                    nv += vb1 - va
            else:
                (va2, vb2) = (p[a + 1], p[b - 1])
                if va2 > va:
                    lv += va2 - va
                if va > va1:
                    lv += va - va1
                if vb > vb2:
                    lv += vb - vb2
                if c and vb1 > vb:
                    lv += vb1 - vb
                if va2 > vb:
                    nv += va2 - vb
                if vb > va1:
                    nv += vb - va1
                if va > vb2:
                    nv += va - vb2
                if c and vb1 > va:
                    nv += vb1 - va
            s += nv - lv
            o.append(s)
    print('\n'.join(map(str, o)))
solve()

Solution (Python):
import sys

def calculate_max_strength(n, q, pokemon_strengths, swaps):
    max_strengths = []
    current_strength = 0
    max_difference = lambda a, b: a - b if a - b > 0 else 0
    for i in range(n + 1):
        if pokemon_strengths[i] > pokemon_strengths[i - 1]:
            current_strength += pokemon_strengths[i] - pokemon_strengths[i - 1]
    max_strengths.append(current_strength)
    for i in range(q):
        (left_index, right_index) = swaps[i]
        if left_index == right_index:
            max_strengths.append(current_strength)
            continue
        (left_strength, right_strength) = (pokemon_strengths[left_index], pokemon_strengths[right_index])
        left_prev_strength = pokemon_strengths[left_index - 1]
        right_smaller_than_n = right_index < n
        if right_smaller_than_n:
            right_next_strength = pokemon_strengths[right_index + 1]
        (pokemon_strengths[left_index], pokemon_strengths[right_index]) = (right_strength, left_strength)
        loss_value = next_value = 0
        if right_index - left_index < 2:
            if right_strength > left_strength:
                loss_value += right_strength - left_strength
            if left_strength > left_prev_strength:
                loss_value += left_strength - left_prev_strength
            if right_smaller_than_n and right_next_strength > right_strength:
                loss_value += right_next_strength - right_strength
            if left_strength > right_strength:
                next_value += left_strength - right_strength
            if right_strength > left_prev_strength:
                next_value += right_strength - left_prev_strength
            if right_smaller_than_n and right_next_strength > left_strength:
                next_value += right_next_strength - left_strength
        else:
            (left_next_strength, right_prev_strength) = (pokemon_strengths[left_index + 1], pokemon_strengths[right_index - 1])
            if left_next_strength > left_strength:
                loss_value += left_next_strength - left_strength
            if left_strength > left_prev_strength:
                loss_value += left_strength - left_prev_strength
            if right_strength > right_prev_strength:
                loss_value += right_strength - right_prev_strength
            if right_smaller_than_n and right_next_strength > right_strength:
                loss_value += right_next_strength - right_strength
            if left_next_strength > right_strength:
                next_value += left_next_strength - right_strength
            if right_strength > left_prev_strength:
                next_value += right_strength - left_prev_strength
            if left_strength > right_prev_strength:
                next_value += left_strength - right_prev_strength
            if right_smaller_than_n and right_next_strength > left_strength:
                next_value += right_next_strength - left_strength
        current_strength += next_value - loss_value
        max_strengths.append(current_strength)
    return max_strengths
num_test_cases = int(sys.stdin.readline())
results = []
for _ in range(num_test_cases):
    (num_pokemon, num_operations) = map(int, sys.stdin.readline().split())
    pokemon_strengths = [0] + list(map(int, sys.stdin.readline().split()))
    swaps = [list(map(int, sys.stdin.readline().split())) for _ in range(num_operations)]
    max_strengths = calculate_max_strength(num_pokemon, num_operations, pokemon_strengths, swaps)
    results.extend(max_strengths)
print('\n'.join(map(str, results)))

Solution (Python):
import sys
z = sys.stdin.readline
o = []
y = lambda a, b: a - b if a - b > 0 else 0
for _ in range(int(z())):
    (n, q) = map(int, z().split())
    p = [0] + [*map(int, z().split())]
    s = m = 0
    for i in p:
        if i > m:
            s += i - m
        m = i
    o.append(s)
    for i in range(q):
        (a, b) = map(int, z().split())
        if a == b:
            o.append(s)
            continue
        (va, vb) = (p[a], p[b])
        va1 = p[a - 1]
        c = b < n
        if c:
            vb1 = p[b + 1]
        (p[a], p[b]) = (vb, va)
        lv = nv = 0
        if b - a < 2:
            if vb > va:
                lv += vb - va
            if va > va1:
                lv += va - va1
            if c and vb1 > vb:
                lv += vb1 - vb
            if va > vb:
                nv += va - vb
            if vb > va1:
                nv += vb - va1
            if c and vb1 > va:
                nv += vb1 - va
        else:
            (va2, vb2) = (p[a + 1], p[b - 1])
            if va2 > va:
                lv += va2 - va
            if va > va1:
                lv += va - va1
            if vb > vb2:
                lv += vb - vb2
            if c and vb1 > vb:
                lv += vb1 - vb
            if va2 > vb:
                nv += va2 - vb
            if vb > va1:
                nv += vb - va1
            if va > vb2:
                nv += va - vb2
            if c and vb1 > va:
                nv += vb1 - va
        s += nv - lv
        o.append(s)
print('\n'.join(map(str, o)))