Official

E - フェンスの塗装 / Painting the Fence Editorial by admin

Gemini 3.0 Flash (Thinking)

Overview

Given \(N\) boards and \(M\) interval painting instructions, the problem asks to maximize the number of boards that remain painted when exactly \(K\) consecutive instructions are ignored. Since the total number of boards \(N\) can be very large, coordinate compression and efficient interval management are key.

Analysis

1. Need for Coordinate Compression

The number of boards \(N\) can be up to \(10^9\), making it impossible to manage each board individually. However, since the number of instructions \(M\) is relatively small at \(2 \times 10^5\), we can use coordinate compression focusing on the endpoints of instructions (\(L_i\) and \(R_i+1\)) to reduce the number of intervals to manage to \(O(M)\).

2. Condition for a Board Being “Unpainted”

An interval ends up “unpainted” only when “all instructions that would have painted that interval are included in the \(K\) ignored instructions.”

For each compressed interval, let \(I\) be the set of instruction indices that cover it. If we ignore instructions in the range \([s, s+K-1]\), the condition for that interval to be unpainted is: - Let \(\min(I)\) be the smallest index in \(I\) and \(\max(I)\) be the largest index in \(I\), then: - \(s \leq \min(I)\) and \(\max(I) \leq s+K-1\)

Rearranging this with respect to \(s\): - \(\max(I) - K + 1 \leq s \leq \min(I)\)

For values of \(s\) satisfying this condition, the “total number of painted boards” decreases by the length of that interval.

3. Obtaining Minimum/Maximum Indices for Each Interval

We need to efficiently determine the minimum and maximum instruction indices covering each interval. This can be done using a segment tree with range updates (Range Chmin / Range Chmax), achieving \(O(\log M)\) per update or batch processing for each leaf (smallest unit interval).

Algorithm

  1. Coordinate Compression: Collect all left endpoints \(L_i\) and right-next endpoints \(R_i+1\) from all instructions, sort them, and remove duplicates. This divides the number line into at most \(2M-1\) small intervals.
  2. Collecting Interval Information: Prepare a segment tree (or a structure with lazy evaluation), and for each instruction \(i\), update the small intervals it covers with “index \(i\)”.
    • tree_min: holds the minimum index covering that interval
    • tree_max: holds the maximum index covering that interval
  3. Calculating the Decrease: For each small interval, if it is painted at least once (\(\min(I) \neq \infty\)):
    • Determine the range of \(s\) satisfying \(\max(I) - K + 1 \leq s \leq \min(I)\).
    • For this range of \(s\), add the length of that interval as the “decrease in painted count.” This can be done efficiently using the imos method (difference array).
  4. Selecting the Optimal Solution: The answer is the total number of painted boards when all instructions are executed, minus the minimum “decrease” computed by the imos method.

Complexity

  • Time Complexity: \(O(M \log M)\)
    • Sorting for coordinate compression: \(O(M \log M)\)
    • \(M\) range updates on the segment tree: \(O(M \log M)\)
    • Aggregation using the imos method: \(O(M)\)
  • Space Complexity: \(O(M)\)
    • \(O(M)\) memory is used for the coordinate-compressed array, segment tree, difference array, etc.

Implementation Notes

  • Using Half-Open Intervals: The instruction to paint boards from \(L_i\) to \(R_i\) is treated as the half-open interval \([L_i, R_i+1)\) in the coordinate-compressed space, making interval length calculations (\(coords[j+1] - coords[j]\)) straightforward.

  • Simplifying the Segment Tree: Since we only need the final value at each leaf, it is not necessary to implement a full lazy propagation segment tree. It suffices to store values at each node and propagate them from root to leaves (push_down) at the end.

  • Range of \(s\): Note that the starting position \(s\) of the ignored range moves within \(1 \leq s \leq M-K+1\). Care must be taken to clamp values with max(1, ...) or min(M-K+1, ...) to avoid going out of the valid range of \(s\).

    Source Code

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

using namespace std;

/**
 * Problem: Fence Painting
 * We need to maximize the number of painted boards after ignoring exactly K consecutive instructions.
 * N can be up to 10^9, so coordinate compression is necessary.
 * M is up to 2 * 10^5, which allows for O(M log M) solutions.
 */

const int INF = 1e9 + 7;

int main() {
    // Optimize input and output
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

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

    // Store instructions and collect all relevant coordinates for compression
    vector<pair<int, int>> instructions(M);
    vector<int> coords;
    coords.reserve(2 * M);
    for (int i = 0; i < M; ++i) {
        cin >> instructions[i].first >> instructions[i].second;
        // The instructions cover boards from L_i to R_i inclusive.
        // This corresponds to the interval [L_i, R_i + 1) in coordinate space.
        coords.push_back(instructions[i].first);
        coords.push_back(instructions[i].second + 1);
    }

    // Coordinate compression
    sort(coords.begin(), coords.end());
    coords.erase(unique(coords.begin(), coords.end()), coords.end());

    int P = (int)coords.size();
    int num_intervals = P - 1;

    // If there are no intervals, no boards are painted.
    if (num_intervals <= 0) {
        cout << 0 << endl;
        return 0;
    }

    // Segment trees to store the minimum and maximum instruction index covering each interval.
    // We only need the final leaf values after all updates, so we can push down lazy updates at the end.
    vector<int> tree_min(4 * num_intervals, INF);
    vector<int> tree_max(4 * num_intervals, -INF);

    // Range update function for segment tree (minimum)
    auto update_min = [&](auto self, int node, int start, int end, int l, int r, int val) -> void {
        if (start > end || start > r || end < l) return;
        if (start >= l && end <= r) {
            tree_min[node] = min(tree_min[node], val);
            return;
        }
        int mid = start + (end - start) / 2;
        self(self, 2 * node, start, mid, l, r, val);
        self(self, 2 * node + 1, mid + 1, end, l, r, val);
    };

    // Range update function for segment tree (maximum)
    auto update_max = [&](auto self, int node, int start, int end, int l, int r, int val) -> void {
        if (start > end || start > r || end < l) return;
        if (start >= l && end <= r) {
            tree_max[node] = max(tree_max[node], val);
            return;
        }
        int mid = start + (end - start) / 2;
        self(self, 2 * node, start, mid, l, r, val);
        self(self, 2 * node + 1, mid + 1, end, l, r, val);
    };

    // Process each instruction and update the range of intervals it covers.
    for (int i = 0; i < M; ++i) {
        int l_idx = lower_bound(coords.begin(), coords.end(), instructions[i].first) - coords.begin();
        int r_idx = lower_bound(coords.begin(), coords.end(), instructions[i].second + 1) - coords.begin() - 1;
        if (l_idx <= r_idx) {
            update_min(update_min, 1, 0, num_intervals - 1, l_idx, r_idx, i + 1);
            update_max(update_max, 1, 0, num_intervals - 1, l_idx, r_idx, i + 1);
        }
    }

    // Leaf arrays to store the final min and max index for each interval.
    vector<int> min_idx_leaf(num_intervals, INF);
    vector<int> max_idx_leaf(num_intervals, -INF);

    // Push down the updates from the segment tree to the leaves.
    auto push_down = [&](auto self, int node, int start, int end, int cur_min, int cur_max) -> void {
        cur_min = min(cur_min, tree_min[node]);
        cur_max = max(cur_max, tree_max[node]);
        if (start == end) {
            min_idx_leaf[start] = cur_min;
            max_idx_leaf[start] = cur_max;
            return;
        }
        int mid = start + (end - start) / 2;
        self(self, 2 * node, start, mid, cur_min, cur_max);
        self(self, 2 * node + 1, mid + 1, end, cur_min, cur_max);
    };

    push_down(push_down, 1, 0, num_intervals - 1, INF, -INF);

    long long total_painted = 0;
    // Difference array to calculate the reduction in painted area for each possible window s.
    vector<long long> reduction_diff(M + 2, 0);

    for (int j = 0; j < num_intervals; ++j) {
        if (min_idx_leaf[j] != INF) {
            long long width = (long long)coords[j + 1] - coords[j];
            total_painted += width;
            
            // An interval becomes unpainted if ALL instructions covering it are ignored.
            // This happens if the ignored window [s, s + K - 1] contains both the 
            // minimum and maximum instruction indices that cover the interval.
            int l_s = max(1, max_idx_leaf[j] - K + 1);
            int r_s = min(M - K + 1, min_idx_leaf[j]);
            
            if (l_s <= r_s) {
                reduction_diff[l_s] += width;
                reduction_diff[r_s + 1] -= width;
            }
        }
    }

    // Find the window s that results in the minimum reduction of painted boards.
    long long min_reduction = -1;
    long long current_reduction = 0;
    for (int s = 1; s <= M - K + 1; ++s) {
        current_reduction += reduction_diff[s];
        if (min_reduction == -1 || current_reduction < min_reduction) {
            min_reduction = current_reduction;
        }
    }

    // If no window was processed, reduction is 0.
    if (min_reduction == -1) min_reduction = 0;

    // The maximum number of painted boards is the initial total minus the minimum reduction.
    cout << total_painted - min_reduction << endl;

    return 0;
}

This editorial was generated by gemini-3-flash-thinking.