Update content/5_Plat/Range_Sweep.mdx

Author: HaccerKat

content/5_Plat/Range_Sweep.mdx

@@ -97,25 +97,25 @@ int main() {
 
 The first step is to create a DP to solve the first subtask. The states are the springboards and the transitions are between springboards. First, sort the springboards by the pair $(x_1, y_1)$ in increasing order. It is possible to show that for all $i$, $j$, where $i < j$, Bessie cannot use springboard $j$ then $i$ later.
 
-For each springboard $i$, let $ans[i]$ denote the minimum distance needed to walk to the start point of springboard $i$.
+For each springboard $i$, let $\texttt{ans}[i]$ denote the minimum distance needed to walk to the start point of springboard $i$.
 
-Let $dist(a, b)$ be the walking distance from the end of springboard $a$ and the start of springboard $b$.
+Let $\texttt{dist}(a, b)$ be the walking distance from the end of springboard $a$ and the start of springboard $b$.
 
-$dist(a, b) = x_1[b] + y_1[b] - x_2[a] - y_2[a]$
+$\texttt{dist}(a, b) = x_1[b] + y_1[b] - x_2[a] - y_2[a]$
 
 Then, the transitions are:
 
-$ans[i] = \min\limits_{j < i, x_2[j] \le x_1[i], y_2[j] \le y_1[i]}(ans[j] + dist(j, i))$ 
+$\texttt{ans}[i] = \min\limits_{j < i, x_2[j] \le x_1[i], y_2[j] \le y_1[i]}(ans[j] + \texttt{dist}(j, i))$ 
 
 **Full Solution:** $\mathcal{O}(P \log P)$
 
 Optimizing the DP involves the use of a min point update range query segment tree. Let's first expand $dist(i, j)$ in the transition formula.
 
-$ans[i] = \min\limits_{j < i, x_2[j] \le x_1[i], y_2[j] \le y_1[i]}(ans[j] + x_1[i] + y_1[i] - x_2[j] - y_2[j])$ 
+$\texttt{ans}[i] = \min\limits_{j < i, x_2[j] \le x_1[i], y_2[j] \le y_1[i]}(\texttt{ans}[j] + x_1[i] + y_1[i] - x_2[j] - y_2[j])$ 
 
-$ans[i] = x_1[i] + y_1[i] + \min\limits_{j < i, x_2[j] \le x_1[i], y_2[j] \le y_1[i]}(ans[j] - x_2[j] - y_2[j])$ 
+$\texttt{ans}[i] = x_1[i] + y_1[i] + \min\limits_{j < i, x_2[j] \le x_1[i], y_2[j] \le y_1[i]}(\texttt{ans}[j] - x_2[j] - y_2[j])$ 
 
-We notice that everything inside the $\min$ statement only depends on $j$. The segment tree stores $ans[j] - x_2[j] - y_2[j]$ at index $y_2[j]$. We can seperate the start and end of springboards to create two seperate events for each springboard, still sorting by $(x, y)$. When the event is the start of a springboard, update $ans[i]$ through a segment tree query. When the event is the end of a springboard, update the segment tree.
+We notice that everything inside the $\min$ statement only depends on $j$. The segment tree stores $\texttt{ans}[j] - x_2[j] - y_2[j]$ at index $y_2[j]$. We can seperate the start and end of springboards to create two seperate events for each springboard, still sorting by $(x, y)$. When the event is the start of a springboard, update $ans[i]$ through a segment tree query. When the event is the end of a springboard, update the segment tree.
 
 By processing in order, the first two conditions in the $\min$ statement are always satisfied. The third is where the segment tree comes into play, where querying the range $[0, y_1[i]]$ is sufficent to satisfy all constraints.