Merge pull request #4557 from brebenelmihnea/maximum-building-2

[Editorial] Maximum Building II

Author: SansPapyrus683

content/4_Gold/Stacks.problems.json

@@ -196,7 +196,7 @@
       "isStarred": false,
       "tags": ["Stack"],
       "solutionMetadata": {
-        "kind": "none"
+        "kind": "internal"
       }
     },
     {

solutions/gold/cses-1148.mdx

@@ -0,0 +1,107 @@
+---
+id: cses-1148
+source: CSES
+title: Maximum Building II
+author: Mihnea Brebenel
+---
+
+## Explanation
+
+Let $r_{i,j}$ be the maximum number of continuous free cells on line $i$ starting from $(i, j)$ to the right.
+This can be precomputed.
+Similarly to [Maximum Building I](https://cses.fi/problemset/task/1147), we can also apply monotonic stacks to find
+1. $u_{i, j}$, the last line above $i$ with cell $(x, j)$ such that $r_{x, j} > r_{i, j}$
+2. $d_{i, j}$, the last line under $i$ with cell $(y, j)$ such that $r_{y, j} \ge r_{i, j}$
+
+We can efficiently update the answer matrix with [prefix sums](/silver/more-prefix-sums#2d-prefix-sums)
+and [difference arrays](https://codeforces.com/blog/entry/78762).
+Having $r_{i,j}$, $u_{i,j}$ and $d_{i,j}$ precomputed, we know the upper bound
+and the lower bound of the rectangle of width $r_{i,j}$,
+i.e. how much it can expand above and below line $i$ maintaing width $r_{i, j}$.
+
+We'll do difference arrays on each column independently.
+Accordingly, the updates of the answer matrix look like this:
+- increment $\texttt{ans}[1][r_{i,j}]$ by $1$
+- decrement $\texttt{ans}[i - u_{i, j} + 2][r_{i,j}]$ by $1$
+- decrement $\texttt{ans}[d_{i,j} - i + 2][r_{i, j}]]$ by $1$
+- increment $\texttt{ans}[d_{i,j} - u_{i,j} + 3][r_{i,j}]$ by $1$
+
+Finally, we add $ans[i][j]$ to $ans[i][j-1]$ i.e. a submatrix of size
+$i \times j$ contains a submatrix of size $i \times (j-1)$.
+
+## Implementation
+
+**Time Complexity:** $\mathcal{O}(N \cdot M)$
+
+<LanguageSection>
+<CPPSection>
+
+```cpp
+#include <iostream>
+#include <stack>
+#include <vector>
+
+using namespace std;
+
+int main() {
+	int n, m;
+	cin >> n >> m;
+	vector<vector<int>> r(n + 2, vector<int>(m + 2));
+	vector<vector<int>> u(n + 1, vector<int>(m + 1));
+	vector<vector<int>> d(n + 1, vector<int>(m + 1));
+	vector<vector<int>> ans(n + 3, vector<int>(m + 3));
+	vector<vector<char>> mat(n + 1, vector<char>(m + 1));
+	for (int i = 1; i <= n; i++) {
+		for (int j = 1; j <= m; j++) { cin >> mat[i][j]; }
+		for (int j = m; j >= 1; j--) {
+			if (mat[i][j] == '*') {
+				r[i][j] = 0;
+			} else {
+				r[i][j] = r[i][j + 1] + 1;
+			}
+		}
+	}
+
+	stack<int> st;
+	// Precompute u[i][j] and d[i][j]
+	for (int j = 1; j <= m; j++) {
+		while (!st.empty()) { st.pop(); }
+		for (int i = 1; i <= n; i++) {
+			while (!st.empty() && r[i][j] < r[st.top()][j]) { st.pop(); }
+			u[i][j] = st.empty() ? 1 : (st.top() + 1);
+			st.push(i);
+		}
+		while (!st.empty()) { st.pop(); }
+		for (int i = n; i >= 1; i--) {
+			while (!st.empty() && r[i][j] <= r[st.top()][j]) { st.pop(); }
+			d[i][j] = st.empty() ? n : (st.top() - 1);
+			st.push(i);
+		}
+	}
+
+	// Make difference array on each column independently
+	for (int i = 1; i <= n; i++) {
+		for (int j = 1; j <= m; j++) {
+			ans[1][r[i][j]]++;
+			ans[i - u[i][j] + 2][r[i][j]]--;
+			ans[d[i][j] - i + 2][r[i][j]]--;
+			ans[d[i][j] - u[i][j] + 3][r[i][j]]++;
+		}
+	}
+	for (int j = 1; j <= m; j++) {
+		for (int i = 1; i <= n; ++i) { ans[i][j] += ans[i - 1][j]; }
+		for (int i = 1; i <= n; ++i) { ans[i][j] += ans[i - 1][j]; }
+	}
+	for (int i = n; i >= 1; i--) {
+		for (int j = m; j >= 2; --j) { ans[i][j - 1] += ans[i][j]; }
+	}
+
+	for (int i = 1; i <= n; i++) {
+		for (int j = 1; j <= m; ++j) { cout << ans[i][j] << " "; }
+		cout << '\n';
+	}
+}
+```
+
+</CPPSection>
+</LanguageSection>