upd

Author: brebenelmihnea

solutions/gold/cses-1148.mdx

@@ -3,8 +3,6 @@ id: cses-1148
 source: CSES
 title: Maximum Building II
 author: Mihnea Brebenel
-prerequisites:
-  - prefix-sums
 ---
 
 ## Explanation
@@ -16,7 +14,7 @@ Similarly to [Maximum Building I](https://cses.fi/problemset/task/1147) apply mo
 
 Using [prefix sums](/silver/more-prefix-sums#2d-prefix-sums) and [difference arrays](https://codeforces.com/blog/entry/78762)
 we can efficiently update the answer matrix. Having $r_{i,j}$, $u_{i,j}$ and $d_{i,j}$ precomputed, we know the upperbound
-and the lowerbound of the rectangle of of width $r_{i,j}$ i.e. how much it can expand above and below line $i$ maintaing width $r_{i, j}$.
+and the lowerbound 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:
 - $ans[1][r_{i,j}]++$
 - $ans[i - u_{i, j} + 2][r_{i,j}]--$ , the upperbound
@@ -58,6 +56,7 @@ int main() {
 		}
 	}
 	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++) {
@@ -72,6 +71,7 @@ int main() {
 			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]]++;