Overview

On an \(H \times W\) grid, road construction (changing empty lots within a rectangular region to roads) is performed \(Q\) times, and after each construction, we need to find the “total number of buildings facing a road.”

Analysis

1. Changes in the Condition for Facing a Road

A building becomes “facing a road” in either of the following cases: - In the initial state, an adjacent cell contains a road. - Due to road construction, an adjacent cell changes from an empty lot to a road.

Once a building faces a road, it remains “facing a road” even if more roads are added by subsequent constructions. Therefore, the approach of sequentially counting “buildings that newly become facing a road” is effective.

2. Area of Impact of Construction

In each construction, all empty lots within the rectangular region \([U, D] \times [L, R]\) become roads. At this point, the only buildings that could potentially newly face a road are buildings immediately adjacent to the outside (top, bottom, left, right) of this rectangular region.

Specifically, it is sufficient to check buildings in the following 4 ranges: 1. One row above the top edge: row \(U-1\), columns \(L \sim R\) 2. One row below the bottom edge: row \(D+1\), columns \(L \sim R\) 3. One column to the left of the left edge: column \(L-1\), rows \(U \sim D\) 4. One column to the right of the right edge: column \(R+1\), rows \(U \sim D\)

Since the constraints state that “the construction range does not contain any buildings,” there is no need to worry about the interior of the rectangle.

3. Complexity Estimation

Naively scanning all cells for each query would result in \(O(Q \times HW)\), which is too slow. However, as described above, if we only check the “boundary of the rectangle,” the processing for each query is proportional to the perimeter of the rectangle. Looking at the constraints, \(\sum (\text{perimeter length}) \leq 5 \times 10^6\), so this computational complexity is well within the time limit.

Algorithm

1. Initial State Counting:
   • Scan all cells, and when a building ‘B’ is found, check whether there is an ‘R’ in the adjacent cells (up, down, left, right).
   • If it faces a road, set the flag is_facing[i][j] to true and increment the current count current_count.
2. Query Processing:
   • For each construction, scan the outside of the 4 edges of the rectangular region.
   • If cell \((i, j)\) is a building ‘B’ and is_facing[i][j] is false:
     • Update is_facing[i][j] to true and increment current_count by 1.
   • At the end of each query, output current_count.

Complexity

• Time Complexity: \(O(HW + \sum_{k=1}^{Q} (\text{Perimeter}_k))\)
  • The initial scan takes \(O(HW)\), and the boundary scan for each query takes \(O(D_k - U_k + R_k - L_k)\). According to the constraints, the total number of operations is approximately \(7 \times 10^6\), which fits within the time limit.
• Space Complexity: \(O(HW)\)
  • \(O(HW)\) memory is used for \(S\) which holds the grid information and the is_facing array (or vector) which holds the state of each building.

Implementation Notes

• Fast I/O: Since \(Q\) and \(HW\) can be large, in C++ it is advisable to use cin.tie(NULL); ios_base::sync_with_stdio(false); to speed up input and output.

• Management with 1D Arrays: Instead of using a 2D array is_facing[H][W], managing it as a 1D array with is_facing[i * W + j] can improve memory efficiency and access speed.

• Preventing Out-of-Bounds Access: When checking the boundary of the rectangle, it is necessary to perform boundary checks with if statements to ensure the boundary does not exceed the edges of the grid (such as row 0 or \(H-1\)).

  Source Code

#include <iostream>
#include <vector>
#include <string>

using namespace std;

/**
 * Problem: City Planning and Road Maintenance
 * We need to count how many buildings 'B' are facing at least one road 'R'.
 * Initially, some buildings might already face roads.
 * Q road construction queries change all 'E' cells in a rectangle [U, D] x [L, R] to 'R'.
 * Buildings are never in the construction region.
 * We need to output the count of buildings facing a road after each query.
 * 
 * Logic:
 * 1. Identify which buildings initially face a road by checking their 4 neighbors.
 * 2. For each query, all cells in the rectangle [U, D] x [L, R] become 'R'.
 * 3. Any building outside the region that is adjacent to any cell inside the region will now face a road.
 * 4. The buildings adjacent to the rectangle are on its four immediate boundaries:
 *    - Row U-1, columns L to R
 *    - Row D+1, columns L to R
 *    - Column L-1, rows U to D
 *    - Column R+1, rows U to D
 * 5. Update the facing status of these buildings and the total count if they weren't already facing a road.
 * 6. Time complexity: O(HW + Sum of Perimeters), which fits within the limits.
 *    Initial check: O(HW), Query processing: O(Sum of (2*(D-U+1) + 2*(R-L+1))).
 *    Given HW <= 2e6 and Sum of Perimeters <= 5e6, total operations are ~7e6.
 */

int main() {
    // Fast I/O for performance
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int H, W, Q;
    if (!(cin >> H >> W >> Q)) return 0;

    // Grid storage: use a vector of strings. HW <= 2e6, so this is memory-efficient.
    vector<string> S(H);
    for (int i = 0; i < H; ++i) {
        cin >> S[i];
    }

    // is_facing[i*W + j] tracks if building (i, j) is facing a road.
    // Using a 1D vector to handle the grid. vector<char> uses 1 byte per element.
    vector<char> is_facing((long long)H * W, 0);
    long long current_count = 0;

    // Initial check for buildings already facing a road in the initial grid.
    for (int i = 0; i < H; ++i) {
        for (int j = 0; j < W; ++j) {
            if (S[i][j] == 'B') {
                bool facing = false;
                // Check 4 cardinal neighbors (Up, Down, Left, Right)
                if (i > 0 && S[i - 1][j] == 'R') facing = true;
                else if (i < H - 1 && S[i + 1][j] == 'R') facing = true;
                else if (j > 0 && S[i][j - 1] == 'R') facing = true;
                else if (j < W - 1 && S[i][j + 1] == 'R') facing = true;
                
                if (facing) {
                    is_facing[(long long)i * W + j] = 1;
                    current_count++;
                }
            }
        }
    }

    // Process Q road construction queries.
    for (int k = 0; k < Q; ++k) {
        int U, D, L, R;
        cin >> U >> D >> L >> R;
        // Convert 1-based coordinates to 0-based.
        --U; --D; --L; --R;

        // Any building adjacent to the query rectangle [U, D] x [L, R]
        // will now face a road. We check the four potential boundary rows/cols.

        // 1. Buildings above the region: row U-1, columns L to R.
        if (U > 0) {
            int i = U - 1;
            for (int j = L; j <= R; ++j) {
                if (S[i][j] == 'B' && !is_facing[(long long)i * W + j]) {
                    is_facing[(long long)i * W + j] = 1;
                    current_count++;
                }
            }
        }
        // 2. Buildings below the region: row D+1, columns L to R.
        if (D < H - 1) {
            int i = D + 1;
            for (int j = L; j <= R; ++j) {
                if (S[i][j] == 'B' && !is_facing[(long long)i * W + j]) {
                    is_facing[(long long)i * W + j] = 1;
                    current_count++;
                }
            }
        }
        // 3. Buildings to the left of the region: column L-1, rows U to D.
        if (L > 0) {
            int j = L - 1;
            for (int i = U; i <= D; ++i) {
                if (S[i][j] == 'B' && !is_facing[(long long)i * W + j]) {
                    is_facing[(long long)i * W + j] = 1;
                    current_count++;
                }
            }
        }
        // 4. Buildings to the right of the region: column R+1, rows U to D.
        if (R < W - 1) {
            int j = R + 1;
            for (int i = U; i <= D; ++i) {
                if (S[i][j] == 'B' && !is_facing[(long long)i * W + j]) {
                    is_facing[(long long)i * W + j] = 1;
                    current_count++;
                }
            }
        }

        // Output the updated count of buildings facing a road after this query.
        cout << current_count << "\n";
    }

    return 0;
}

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