For the interested reader, this problem draws inspiration from the Secretary Problem. In terms of this problem, each one of Farmer's John's memories can be viewed as the following: If Farmer John were to uniformly reject the first $a_i$ cows, and then choose the first cow he saw that was better than any cow he had seen before, he would choose the $h_i$-th cow.

Subtask 1:

We can solve this with brute force. Simply iterate through all possible sequences of cowmpetency scores and return the first one that satisfies Farmer John's memories.

Code:

def genarray(i, N, C, curscores):
    if curscores[i] != 0:
        if i == N: yield curscores
        else: yield from genarray(i + 1, N, C, curscores)
    else:
        for score in range(1, C + 1):
            curscores[i] = score
            if i == N: yield curscores
            else: yield from genarray(i + 1, N, C, curscores)
        curscores[i] = 0

t = int(input())

for test in range(t):
    N, Q, C = map(int, input().strip().split())
    c = [0] + [*map(int, input().strip().split())]
    constraints = [[*map(int, input().strip().split())] for _ in range(Q)]
    for curscores in genarray(1, N, C, c):
        for [a, h] in constraints:
            if max(curscores[: a + 1]) != max(curscores[:h]) or \
                curscores[h] <= max(curscores[: a + 1]):
                break
        else:
            print(*curscores[1:])
            break
    else:
        print(-1)

Subtask 2:

Let us now define $B(k)$ to be the index of the first cow with a strictly larger score than the first $k$ cows. This is useful because each of Farmer John's memories can be fully described by the statement: $B(a_i) = h_i$.

Based on this and the above brute force, we realize that possible sequences only satisfy Farmer John's constraints if, for every $1 \leq i \leq N$:

1. $\max\limits_{1 \leq j \leq i}{c_j} = \max\limits_{1 \leq j < B(i)}{c_j}$
2. $c_{B(i)} > \max\limits_{1 \leq j \leq i} c_j$

This leads us to the following observation: For all $1 \leq i < j < B(i)$: $B(i) = B(j)$. Therefore, if there exists $i,j$ such that $a_i < a_j < B(a_i)$ and $B(a_i) \neq B(a_j)$ then there does not exist a sequence in accordance with Farmer John's memory. Combining these observation, with the given $(a_i, h_i)$ pairs, we can derive the value $B(j)$ for any $j$ that satisfies the inequality $a_i \leq j < h_i$ for some $i$. For all the other values of $j$ we will call $B(j)$ unknown

This is enough information to come up with a greedy solution. We will iterate through $i$ from $1$ to $N$. At each point, we will store the minimum possible values $c_1 \dots c_n$ that satisfy the constraints related to $B(1) \dots B(i)$. This means that all values of $c_i$ that were not given are initially set to $1$ since this is the minimum possible.

At a certain $i$, if $B(i)$ is unknown then no change will be made to the array since there are no associated constraints with $B(i)$. Otherwise, there are two cases:

1. $\max\limits_{1 \leq j \leq i}{c_j} = \max\limits_{1 \leq j < B(i)}{c_j}$: We can see that this applies no additional constraint to $c_i$ so we do not have to change it. There is an associated change with $c_{B(i)}$ which we'll discuss after the next case.
2. Otherwise we have: $\max\limits_{1 \leq j \leq i}{c_j} < \max\limits_{1 \leq j < B(i)}{c_j}$. In this case we need to find a $j \leq i$ and set $c_j$ to $\max\limits_{1 \leq \ell < B(i)}{c_\ell}$. $j$ needs to be as large as possible such that $c_j$ was not given in the problem and there exists no $k$ such that $k \leq j$ and $B(k) < B(i)$. The reasoning behind the first part is simple, we're trying to find the lexographically minimum sequence. So if we have to increase some $c_j$, we want to pick as large of a $j$ as possible. The reason for the second part is the following: if we were to change $c_j$ and such a $k$ existed, we would have to satisfy $\max\limits_{1 \leq \ell < B(i)}{c_\ell} = c_j < c_{B(k)} \leq \max\limits_{1 \leq \ell < B(i)}{c_\ell}$. This is clearly not possible. Therefore, if no such $j$ exists, a sequence is impossible to construct.

For both of these cases, we also need to ensure that $c_{B(i)} > \max\limits_{1 \leq j \leq i} c_j$. If $c_{B(i)}$ is not given, then we can set $c_{B(i)} = \max\limits_{1 \leq j \leq i} c_j + 1$. This is the smallest possible value. If $c_{B(i)}$ was given, we need to make sure that it is at least this value, otherwise, no sequence is possible.

The reason this algorithm works is that each additional constraint that is added can only result in an equivalent or larger lexicographic sequence. Therefore, by making changes as far back in the array as possible, the resulting sequence is the lexicographically minimum one.

To finish, it's now simple to iterate through the sequence and ensure that all the values are at most $C$ to check if the last constraint in the problem is satisfied.

t = int(input())

def solve():
    N, Q, C = map(int, input().strip().split())
    B = [0] * (N + 1)
    c = [0] + [*map(int, input().strip().split())]
    assigned = [*map(bool, c)]

    # Initially choose 1 for every non-fixed element
    c = [score if score else 1 for score in c]

    for _ in range(Q):
        ai, hi = map(int, input().strip().split())
        B[ai] = hi

    # Ensure constraints for B
    for i in range(1, N + 1):
        # Ensure that every j such that i < j < B[i] has B[j] = B[i]
        for j in range(i, B[i]):
            if B[j] != 0 and B[j] != B[i]:
                return print(-1)
            B[j] = B[i]

    for i in range(1, N + 1):
        if B[i] == 0:
            continue
        # calculate max before and max after
        mx_before = max(c[: i + 1])
        mx_after = max(c[: B[i]])

        # change max before such that mx_before = mx_after
        if mx_after > mx_before:
            for j in range(i, 0, -1):
                if B[j] != 0 and B[j] < B[i]:
                    return print(-1)
                if assigned[j]:
                    continue
                c[j] = mx_after
                break
            else:
                return print(-1)
            mx_before = mx_after

        if not assigned[B[i]]:
            c[B[i]] = mx_before + 1
        # check to make sure B[i] > mx_before
        if c[B[i]] <= mx_before:
            return print(-1)

    # Check that each element in the minimum sequence is at most C
    for i in range(1, N + 1):
        if c[i] > C:
            return print(-1)

    return print(*c[1:])


for _ in range(t):
    solve()

Subtask 3 In this case, we have to optimize the two $O(n^2)$ parts of our solution. The first part, filling in known values of $B$ can be solved using two pointers. To optimize the greedy algorithm we can notice that all values $j$ where $i \leq j < B(i)$ satisfy $B(i) = B(j)$. Therefore, once you satisfy the constraint associated with $B(i)$, the ones associated with $B(i+1) \dots B(B(i)-1)$ are redundant so we can skip them. Putting this together, we get an algorithm that runs in $O(n)$.

Code:

t = int(input())

def solve():
    N, Q, C = map(int, input().strip().split())
    B = [0] * (N + 1)
    c = [0] + [*map(int, input().strip().split())]
    assigned = [*map(bool, c)]

    # Initially choose 1 for every non-fixed element
    c = [score if score else 1 for score in c]

    for _ in range(Q):
        ai, hi = map(int, input().strip().split())
        B[ai] = hi

    # Ensure constraints for B
    curind = 1
    while curind <= N:
        i = curind
        # Ensure that every j such that i < j < B[i] has B[j] = B[i]
        while curind < B[i]:
            if B[curind] != 0 and B[curind] != B[i]:
                return print(-1)
            B[curind] = B[i]
            curind += 1
        curind = max(curind, i+1)

    mx_before = 0
    mx_after = 0
    i = 1
    while i <= N:
        # calculate max before and max after
        mx_before = max(mx_before, c[i])
        mx_after = max(mx_after, c[i])
        if B[i] == 0:
            i += 1
            continue
        mx_after = max(mx_after, *c[i:B[i]])
        
        # change mx_before such that mx_before = mx_after
        if mx_after > mx_before:
            for j in range(i, 0, -1):
                if B[j] != 0 and B[j] < B[i]:
                    return print(-1)
                if assigned[j]:
                    continue
                c[j] = mx_after
                break
            else:
                return print(-1)
            mx_before = mx_after

        if not assigned[B[i]]:
            c[B[i]] = mx_before + 1
        # check to make sure B[i] > mx_before
        if c[B[i]] <= mx_before:
            return print(-1)

        i = B[i]

    # Check that each element in the minimum sequence is at most C
    for i in range(1, N + 1):
        if c[i] > C:
            return print(-1)

    return print(*c[1:])

for _ in range(t):
    solve()