Overview

Since the start position of the inspection interval is an integer, there are a finite number of candidates: \(0, 1, \dots, T-K\).
We manage which event venues overlap with each start position using bitsets, and test “sets of events that are allowed to be affected” in increasing order of size to determine if the interval constraints can be satisfied.

Analysis

When placing the inspection interval \((s,s+K)\), the condition for it to overlap with the event venue \((L_i,R_i)\) is

\[ (s,s+K) \cap (L_i,R_i) \neq \emptyset \]

Since these are open intervals, the condition for them to overlap is

\[ s < R_i \quad \text{and} \quad L_i < s+K \]

Since \(s\) is an integer, this is equivalent to

\[ s \leq R_i-1 \]

and

\[ s \geq L_i-K+1 \]

Therefore, we can precompute the “set of events that overlap with the inspection interval” for each start position \(s\).

---

Next, we want to minimize the number of affected event venues.

Since \(N \leq 12\) is small, we can represent the set of events as a bitmask.

For example, let a set \(A\) be the “set of events that are allowed to be affected”.
In this case, the start positions \(s\) we are allowed to use are only those where:

• The inspection interval \((s,s+K)\) does not overlap with any events not included in \(A\).

In other words, the problem reduces to the following decision problem:

If we only allow events in the set \(A\) to be affected, can we place the inspection intervals to satisfy the interval constraints?

We can test this for \(|A|=0,1,2,\dots,N\) in increasing order, and the first size for which it is possible will be the answer.

---

Naively searching all configurations of inspection intervals would result in a very large number of combinations, as there are up to \(T-K+1\) start positions and we can choose any number of them.

Therefore, for a fixed allowed set \(A\), we use a greedy approach to quickly determine whether a valid placement is possible.

---

Let’s rewrite the interval constraints using only the start positions.

Let the start positions of the inspection intervals in ascending order be \(s_1,s_2,\dots,s_p\). For \(p \geq 1\):

• For the first inspection interval:

\[ s_1 \leq M \]

• For adjacent inspection intervals:

\[ s_{j+1} - (s_j+K) \leq M \]

which simplifies to:

\[ s_{j+1} \leq s_j + K + M \]

• For the last inspection interval:

\[ T - (s_p+K) \leq M \]

which simplifies to:

\[ s_p \geq T-K-M \]

Therefore, for a fixed allowed set \(A\), the problem becomes:

• Only consider the valid start positions.
• We can start at any position less than or equal to \(M\).
• The next start position can be connected if it is within \(K+M\) of the previous start position.
• We succeed if we can reach a start position greater than or equal to \(T-K-M\).

Algorithm

First, if \(T \leq M\), we can place \(0\) inspection intervals.
In this case, \(0\) event venues are affected, so the answer is \(0\).

From now on, we assume \(T > M\).

---

1. Precompute the set of overlapping events for each start position

The start position \(s\) of the inspection interval satisfies:

\[ 0 \leq s \leq T-K \]

For each \(s\), we record the overlapping event venues in a bitmask start_masks[s].

The range of \(s\) that overlaps with event \(i\) is:

\[ L_i-K+1 \leq s \leq R_i-1 \]

However, since valid start positions must satisfy \(0 \leq s \leq T-K\), we clamp the range to this interval.

---

2. Test the allowed event sets in increasing order of size

We represent the set of events as a bitmask.

Let allowed be the “set of events that are allowed to be affected”.
If we let forbidden be its complement, the condition for the start position \(s\) to be usable is:

start_masks[s] & forbidden == 0

This means “it does not overlap with any forbidden events”.

---

3. Greedily determine reachability for a fixed allowed

We define:

\[ D = K+M \]

This is the maximum distance allowed to the next start position.

Also, the start position of the last inspection interval must satisfy:

\[ s \geq T-K-M \]

In the code, this is represented as need_last.

---

We iterate through the usable start positions in ascending order.

Let cur be the “rightmost start position that is currently reachable”.

When a start position \(s\) is usable:

• If \(s \leq M\), it can be placed as the first inspection interval.
• If \(s \leq cur + K + M\), it can be placed next to the current configuration.

Thus, such an \(s\) is reachable.

If it is reachable, we update cur = s.

Since we are iterating in ascending order, it is sufficient to keep track of only the rightmost reachable start position. This is because reaching a position further to the right expands the range where the next interval can be placed.

If at any point:

\[ cur \geq T-K-M \]

then the final condition is also satisfied, meaning placement is possible with the given allowed set.

---

4. Output the first successful set size

Since we test the size of allowed in the order \(0,1,2,\dots,N\), the size at which placement first becomes possible is the minimum number of affected event venues.

Complexity

Let \(S=T-K+1\) be the number of start positions.

• Precomputation of event sets for each start position: \(O(NS)\)
• Enumeration of event sets: at most \(2^N\)
• Greedy check for each set: \(O(S)\)

Therefore:

• Time Complexity: \(O(NS + 2^N S)\)
• Space Complexity: \(O(S + 2^N)\)

Given the constraints \(N \leq 12\) and \(T \leq 1000\), this is sufficiently fast.

Implementation Points

Be careful with the overlap check of open intervals.

The condition for \((s,s+K)\) and \((L_i,R_i)\) to overlap is:

\[ s < R_i \]

and

\[ L_i < s+K \]

Since \(s\) is an integer, in the code we find the range of start positions overlapping with event \(i\) as follows:

lo = max(0, L - K + 1)
hi = min(max_s, R - 1)

Also, if \(T \leq M\), we have the option of placing zero inspection intervals, so we handle this as a special case at the beginning and immediately output 0.

Source Code

import sys

def main():
    data = list(map(int, sys.stdin.buffer.read().split()))
    if not data:
        return

    T, N, K, M = data[0], data[1], data[2], data[3]
    idx = 4

    events = []
    for _ in range(N):
        L, R = data[idx], data[idx + 1]
        idx += 2
        events.append((L, R))

    if T <= M:
        print(0)
        return

    max_s = T - K
    start_masks = [0] * (max_s + 1)

    for i, (L, R) in enumerate(events):
        bit = 1 << i
        lo = max(0, L - K + 1)
        hi = min(max_s, R - 1)
        if lo <= hi:
            for s in range(lo, hi + 1):
                start_masks[s] |= bit

    total_masks = 1 << N
    full_mask = total_masks - 1

    popcount = [0] * total_masks
    buckets = [[] for _ in range(N + 1)]
    for mask in range(total_masks):
        if mask:
            popcount[mask] = popcount[mask >> 1] + (mask & 1)
        buckets[popcount[mask]].append(mask)

    D = K + M
    need_last = T - K - M

    for ans in range(N + 1):
        for allowed in buckets[ans]:
            forbidden = full_mask ^ allowed
            cur = -10**18

            for s, sm in enumerate(start_masks):
                if sm & forbidden:
                    continue

                if s <= M or s <= cur + D:
                    cur = s
                    if s >= need_last:
                        print(ans)
                        return

if __name__ == "__main__":
    main()

---

This editorial was generated by gpt-5.5-high.