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# Sparse Segment Trees

Author: Andi Qu

### Prerequisites

Querying big ranges.

ResourcesSolutionProblems
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Focus Problem – read through this problem before continuing!

In problems where the query range is at most something like $10^6$, a normal segment tree suffices. However, as soon as we move to bigger ranges ($10^{12}$ in some cases), a normal segment tree results in MLE.

Luckily, we can still use a segment tree to solve these types of problems.

## Resources

The main idea is that we don't have to store all the nodes at all times - we create nodes only when needed. In a normal segment tree, an update only affects $O(\log N)$ nodes, so in a sparse segment tree, we only store $O(Q \log N)$ nodes!

We can implement this efficiently using pointers to a node's children - just like a trie! (Then again, a segment tree is basically a fancier binary trie.)

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## Solution

We need to support two operations on a range:

• Count the number of red-ripe trees in a range (range sum)
• Set all trees in a range to red-ripe (range paint)

We can use a segment tree with lazy propagation to solve this, but the query range is up to $10^9$, so we have to use a sparse segment tree.

Luckily, lazy propagation still works here.

C++

1#include <bits/stdc++.h>2#pragma GCC optimize("O3")3#define FOR(i, x, y) for (int i = x; i < y; i++)4#define MOD 10000000075typedef long long ll;6using namespace std;7
8struct Node {9    int sum, lazy, tl, tr, l, r;10    Node() : sum(0), lazy(0), l(-1), r(-1) {}
Optional

It's possible to reduce the memory of a sparse segment tree to $O(N)$, as described here.

### Problems

StatusSourceProblem NameDifficultyTagsSolution
IOINormalExternal Sol
Balkan OINormal