usaco-guide/content/5_Gold/Springboards.md

id	title	author	prerequisites	description
springboards	Max Suffix Query with Insertions Only	Benjamin Qi	Silver - More with Maps & Sets Silver - Amortized Analysis	Solving USACO Gold - Springboards.

To solve USACO Gold Springboards, we need a data structure that supports operations similar to the following:

1. Add a pair (a,b).
2. For any x, query the maximum value of b over all pairs satisfying a\ge x.

This can be solved with a segment tree (see "Point Update Range Sum"), but a simpler option is to use a map. We rely on the fact that if there exist pairs (a,b) and (c,d) in the map such that a\le c and b\le d, we can simply ignore (a,b) (and the answers to future queries will not be affected). So at every point in time, the pairs (a_1,b_1),(a_2,b_2),\ldots,(a_k,b_k) that we store in the map will satisfy a_1 < a_2 < \cdots < a_k and b_1 > b_2 > \cdots > b_k.

• Querying for a certain x can be done with a single lower_bound operation, as we just want the minimum i such that a_i\ge x.
• When adding a pair (a',b'), first check if there exists (a,b) already in the map such that a\ge a', b\ge b'.
  • If so, then do nothing.
  • Otherwise, insert (a',b') into the map and repeatedly delete pairs (a,b) such that a\le a', b\le b' from the map until none remain.

If there are N insertions, then each query takes O(\log N) time and adding a pair takes O(\log N) time amortized.

#define f first
#define s second
#define lb lower_bound

map<int,ll> m;
void ins(int a, ll b) {
	auto it = m.lb(a); if (it != end(m) && it->s >= b) return;
	it = m.insert(it,{a,b}); it->s = b;
	while (it != begin(m) && prev(it)->s <= b) m.erase(prev(it));
}
ll query(int x) { auto it = m.lb(x); 
	return it == end(m) ? 0 : it->s; } 
// it = end(m) means that no pair satisfies a >= x

Problems

(easier examples?)

- For each $a$ from $p\to 1$, calculate the number of possible cards with that value of $a$. - Genfuncs not required but possibly helpful