How to Cover an N × N Square Room with the least Square Tiles? (could be reused)

Summary

The problem of covering an N × N square room with square tiles without overlapping can be solved using dynamic programming and geometry. The goal is to find the minimum number of tiles required to cover the room, where the side length of each tile is less than N.

Root Cause

The root cause of the complexity in this problem lies in the fact that the tiles do not all need to be the same size, and the side length of each tile is less than N. This introduces a large number of possible combinations, making it challenging to find the optimal solution.

Why This Happens in Real Systems

In real-world systems, this problem can occur in various scenarios, such as tiling a floor or a wall with square tiles. The goal is to minimize the number of tiles used while ensuring that the entire area is covered without overlapping.

Real-World Impact

The real-world impact of this problem is significant, as it can result in cost savings and improved efficiency in various industries, such as construction and manufacturing. By finding the minimum number of tiles required, companies can reduce waste and optimize their production processes.

Example or Code

#include 
using namespace std;

const int inf = 0x3f3f3f3f;
int vis[105][105];
int n;

bool judge(int i, int j, int k) {
    bool ans = true;
    for (int ii = i; ii < i + k && ans; ii++) {
        for (int jj = j; jj < j + k && ans; jj++) {
            if (vis[ii][jj] == true) ans = false;
        }
    }
    return ans;
}

void change(bool x, int i, int j, int k) {
    for (int ii = i; ii < i + k; ii++) {
        for (int jj = j; jj < j + k; jj++) {
            vis[ii][jj] = x;
        }
    }
    return;
}

int dfs(int i, int j) {
    int ans = 0;
    int cur = 0;
    if (i == n && j == n) {
        if (vis[i][j] == false) return 1;
        else return 0;
    }
    if (vis[i][j] == true) {
        if (j < n) ans += dfs(i, j + 1);
        else ans += dfs(i + 1, 1);
    } else {
        ans = inf;
        if (i == 1 && j == 1) {
            for (int k = 1; k <= n / 2; k++) {
                cur = 0;
                if (judge(i, j, k)) {
                    cur++;
                    change(true, i, j, k);
                    if (j < n) cur += dfs(i, j + 1);
                    else cur += dfs(i + 1, 1);
                    change(false, i, j, k);
                    ans = min(ans, cur);
                }
            }
        } else
            for (int k = 1; k < n; k++) {
                cur = 0;
                if (judge(i, j, k)) {
                    cur++;
                    change(true, i, j, k);
                    if (j < n) cur += dfs(i, j + 1);
                    else cur += dfs(i + 1, 1);
                    change(false, i, j, k);
                    ans = min(ans, cur);
                }
            }
    }
    return ans;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        cin >> n;
        for (int i = 0; i <= n + 1; i++) {
            for (int j = 0; j <= n + 1; j++) {
                if (i == 0 || i == n + 1 || j == 0 || j == n + 1) vis[i][j] = true;
                else vis[i][j] = false;
            }
        }
        cout << dfs(1, 1) << endl;
    }
}

How Senior Engineers Fix It

Senior engineers can fix this problem by using dynamic programming to store the minimum number of tiles required for each sub-problem. They can also use geometry to identify the optimal tile sizes and positions.

Why Juniors Miss It

Juniors may miss this problem because they may not have enough experience with dynamic programming and geometry. They may also overlook the fact that the tiles do not all need to be the same size, which introduces additional complexity to the problem.