Overview

This problem asks us to optimally place several inspection intervals (each of length \(K\)) along a road of total length \(T\) km. While satisfying the interval constraints, we want to minimize the number of inspection intervals that overlap with the \(N\) given event venues (represented as open intervals).

By noting that the number of event venues \(N\) is very small (at most \(12\)), we can efficiently find the correct answer within the time limit by combining bitmask brute-force (\(2^N\) possibilities) and dynamic programming (DP).

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Analysis

1. Condition for “Overlap” between Event Venues and Inspection Intervals

Let \(s\) be the starting point of an inspection interval. This interval can be represented as the open interval \((s, s+K)\). We consider the condition under which this interval overlaps (has a non-empty intersection) with the \(i\)-th event venue \((L_i, R_i)\).

The condition for two open intervals to overlap is that the following inequalities hold simultaneously: $\(s < R_i \quad \text{and} \quad L_i < s + K\)$

Rearranging this for \(s\) yields: $\(L_i - K < s < R_i\)$

Since \(s\) is an integer, this condition is equivalent to: $\(L_i - K + 1 \leq s \leq R_i - 1\)$

In other words, if we place an inspection interval starting at \(s\) within this range, it will overlap with event venue \(i\).

2. Utilizing the Constraint \(N \leq 12\)

Since the number of event venues \(N\) is extremely small, we can brute-force all combinations of “which event venues to avoid (i.e., not overlap with)”. Representing the set of avoided event venues as a bitmask (mask), there are only \(2^N = 2^{12} = 4096\) possible combinations.

If the number of avoided event venues is \(d\) (the number of set bits in mask), then the number of overlapping event venues is \(N - d\). Since we want to minimize the number of overlaps, we can search the masks in descending order of the number of avoided venues \(d\) (from \(N\) down to \(0\)). For each mask, we check whether there exists a valid placement of inspection intervals that satisfies the interval constraints while avoiding all event venues included in the mask.

3. Determining Valid Placements under Interval Constraints (DP)

Given a specific mask (the set of event venues to avoid), we determine whether we can place the inspection intervals without overlapping with them.

First, for each event venue \(i\) that must be avoided, we set the forbidden area for the starting point as \([L_i - K + 1, R_i - 1]\). This determines whether it is allowed to place an interval starting at each \(s \in [0, T-K]\) (allowed[s]).

Next, we use dynamic programming (DP) to determine whether we can choose the starting points of the inspection intervals \(s_1 < s_2 < \dots < s_p\) such that they satisfy the interval constraints.

• \(dp[x]\): whether there exists a valid placement satisfying the interval constraints such that an inspection interval starts at \(x\) (true / false).

Initialization

The starting point of the first inspection interval, \(s_1\), must satisfy \(s_1 \leq M\). Therefore, for all \(x\) such that \(0 \leq x \leq \min(T-K, M)\) and allowed[x] is true, we set \(dp[x] = \text{true}\).

Transitions

The distance between the starting points of adjacent inspection intervals must satisfy \(s_{j+1} - s_j \leq K + M\). Therefore, if there exists some \(y\) such that \(dp[y] = \text{true}\), \(x - y \leq K + M\), and allowed[x] is true, we can update \(dp[x] = \text{true}\).

A naive implementation of this would take \(O(T^2)\) time. However, by maintaining the index of the most recent true value (last_true), we can update the DP in \(O(T)\) time while scanning \(x\) from left to right.

Goal Condition

The starting point of the last inspection interval, \(s_p\), must satisfy \(T - (s_p + K) \leq M \iff s_p \geq T - K - M\). After completing the DP, if there exists at least one \(y\) in the range \(T - K - M \leq y \leq T - K\) such that \(dp[y] = \text{true}\), we can conclude that a valid placement exists.

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Algorithm

1. Handling Corner Cases: If \(T \leq M\), we can place \(0\) inspection intervals (resulting in \(0\) overlaps). Thus, we can immediately print 0 and terminate.
2. Preparing the Search Order: Group the \(2^N\) possible masks in descending order of the number of set bits (popcount).
3. Bitmask Brute-force and DP: For each number of avoided venues \(d = N, N-1, \dots, 0\) in descending order, we perform the following check (solve(mask)) for each corresponding mask:
   • Forbid any starting point \(s\) that overlaps with the event venues specified by mask.
   • Initialize the DP table and determine if a placement is possible at each point using \(O(T)\) transitions.
   • If a placement satisfying the goal condition exists, terminate the search immediately, print \(N - d\), and exit.

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Complexity

Time Complexity: \(O(2^N \cdot (N + T))\)

• There are \(2^N\) possible masks.
• In the check process (solve) for each mask:
  • Setting forbidden intervals: If the number of avoided event venues is \(d\), filling the forbidden intervals for each venue takes \(O(d \cdot T)\).
  • Updating the DP table and checking the goal condition: \(O(T)\).
• Even in the worst case, the total number of steps is around \(2^{12} \times 12 \times 1000 \approx 4.9 \times 10^7\) simple loop iterations, which is fast enough to easily run within the time limit (taking only a few to tens of milliseconds).

Space Complexity: \(O(2^N + T)\)

• We store \(2^N\) elements in masks_by_pop.
• The sizes of the DP table and the allowed array are \(O(T)\).
• The memory usage is around a few megabytes, which is well within the limit.

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Key Implementation Points

• Handling Boundary Values of Open Intervals: To avoid overlaps between inspection intervals and event venues, the indices of the forbidden interval \(L_i - K + 1 \leq s \leq R_i - 1\) are properly clipped using max(0, ...) and min(max_s, ...) so that they do not exceed the road boundaries \([0, T-K]\).

• Optimizing the DP: By using \(O(1)\) transitions with last_true, the inner DP loop is processed extremely fast.

• Optimization via Early Termination: Since we search in descending order of the number of avoided event venues \(d\) (ascending order of the number of overlapping venues), the first valid placement found is guaranteed to be the globally optimal solution, allowing us to terminate the program immediately.

  Source Code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    // 高速な入出力
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int T, N, K, M;
    if (!(cin >> T >> N >> K >> M)) return 0;

    vector<int> L(N), R(N);
    for (int i = 0; i < N; ++i) {
        cin >> L[i] >> R[i];
    }

    // p = 0 の場合、点検区間を 0 個設置して T <= M を満たせば重なりは 0
    if (T <= M) {
        cout << 0 << "\n";
        return 0;
    }

    // p >= 1 の場合
    int max_s = T - K;
    vector<bool> allowed(max_s + 1);
    vector<bool> dp(max_s + 1);

    // mask で指定されたイベント会場と重ならないように配置できるか判定
    auto solve = [&](int mask) -> bool {
        fill(allowed.begin(), allowed.end(), true);
        for (int i = 0; i < N; ++i) {
            if ((mask >> i) & 1) {
                // 開区間 (L_i - K, R_i) に含まれる整数 s を禁止する
                int left = max(0, L[i] - K + 1);
                int right = min(max_s, R[i] - 1);
                for (int x = left; x <= right; ++x) {
                    allowed[x] = false;
                }
            }
        }

        fill(dp.begin(), dp.end(), false);
        // 初期化: s_1 <= M かつ s_1 は許容領域内
        for (int x = 0; x <= min(max_s, M); ++x) {
            if (allowed[x]) {
                dp[x] = true;
            }
        }

        int last_true = -1;
        for (int x = 0; x <= max_s; ++x) {
            if (dp[x]) {
                last_true = x;
            } else if (allowed[x]) {
                if (last_true != -1 && x - last_true <= K + M) {
                    dp[x] = true;
                    last_true = x;
                }
            }
        }

        // ゴール判定: T - (s_p + K) <= M  <=>  s_p >= T - K - M
        for (int y = max(0, max_s - M); y <= max_s; ++y) {
            if (dp[y]) {
                return true;
            }
        }
        return false;
    };

    // 重ならないイベント会場の数を最大化するため、popcount の降順に探索
    vector<vector<int>> masks_by_pop(N + 1);
    for (int mask = 0; mask < (1 << N); ++mask) {
        int pop = __builtin_popcount(mask);
        masks_by_pop[pop].push_back(mask);
    }

    for (int d = N; d >= 0; --d) {
        for (int mask : masks_by_pop[d]) {
            if (solve(mask)) {
                cout << N - d << "\n";
                return 0;
            }
        }
    }

    return 0;
}

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