Sparse Segment Trees

Author: Andi Qu

Querying big ranges.

Focus Problem – read through this problem before continuing!

In problems where the query range is at most something like 10610^6, a normal segment tree suffices. However, as soon as we move to bigger ranges (101210^{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.


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(logN)\mathcal{O}(\log N) nodes, so in a sparse segment tree, we only store O(QlogN)\mathcal{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.)

This section is not complete.

Any help would be appreciated! Just submit a Pull Request on Github.


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 10910^9, so we have to use a sparse segment tree.

Luckily, lazy propagation still works here.


#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#define FOR(i, x, y) for (int i = x; i < y; i++)
#define MOD 1000000007
typedef long long ll;
using namespace std;
struct Node {
int sum, lazy, tl, tr, l, r;
Node() : sum(0), lazy(0), l(-1), r(-1) {}


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


Some of these problems are solvable with a normal segment tree if you use coordinate compression. Sparse segment trees are rarely ever required to solve a particular problem, but they can make things a lot more convenient.

StatusSourceProblem NameDifficultyTags
Balkan OINormal

Module Progress:

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