Li Chao Tree is a data structure that offers a method to insert specific functions and query the minimum (or maximum) value at a point in logarithmic time. It was first introduced by Li Chao during a lecture in 2013.

For any pair of inserted functions,

KaTeX can only parse string typed expression

f_{i} (x)

and

f_{j} (x)

(where

1 \leq i < j \leq n

), there exists a point

p

such that:

$f_{i} (x) \leq f_{j} (x)$ when $x \leq p$ , and
$f_{i} (x) \geq f_{j} (x)$ when $x \geq p$

(or the conditions could be in reverse order) for all

KaTeX can only parse string typed expression

x

in the possible query range. The concept behind the Li Chao Tree bears some resemblance to the Segment Tree, but what distinguishes it is the unique way it stores function information in each node.

Motivational Problem

You are given

KaTeX can only parse string typed expression

n

functions

f_{i} (x) = a_{i} x^{3} + b_{i} x^{2} + c_{i} x + d_{i}

(where

1 \leq i \leq n

). For each of the

q

queries, determine

min_{1 \leq i \leq n} f_{i} (x_{j})

for a given

x_{j}

Constraints given:

$1 \leq n, q \leq 1 0^{5}$
$0 \leq a_{i} \leq 1 0^{3}$ (for each $i$ such that $1 \leq i \leq n$ )
$0 \leq b_{i}, c_{i}, d_{i} \leq 1 0^{5}$ (for each $i$ such that $1 \leq i \leq n$ )
$0 \leq x_{j} \leq 1 0^{5}$ (for each $j$ such that $1 \leq j \leq q$ )

This problem comes straight from CodeChef - Polynomials.

Li Chao Tree

Much like the Segment Tree, we want to perform a traversal of nodes from root to leaf. Each node is expected to store a function information such that when we traverse from root to leaf, it is guaranteed that one of the functions met in the path gives the minimum (or maximum) value. For the sake of simplicity, let's assume that our problem queries for minimum value.

Suppose we want to insert a function

KaTeX can only parse string typed expression

f_{j} (x)

. Denote

g (x)

as a function that were saved in the current node resulting from previous insertions. Define

mid

as the middle-point of the range, e.g. a segment

[L, R)

, in the current node. Then, we may perform two comparisons:

f_{j} (L)

with

g (L)

and

f_{j} (mid)

with

g (mid)

Case
KaTeX can only parse string typed expression
fj(L)≥g(L) and $f_{j} (mid) \geq g (mid)$ :

This implies that there are three possibilities: there is no
KaTeX can only parse string typed expression
p or $p < L$ or $p \geq mid$ .
- there is no $p$ that meets our criteria; hence, no action is required.
- $p < L$ , and again, no action is needed as it is outside our considered interval.
- $p \geq mid$ , which means $p$ may affect the second half of the current segment, i.e. $[mid, R)$ . Thus, we need to recurse to the right child of the current node to handle that case.
We also don't need to update the current node as
KaTeX can only parse string typed expression
g(x)≤fj(x) when $L \leq x \leq mid$ . Whenever a query of $L \leq x \leq mid$ is performed, the traversal from root to leaf is guaranteed to pass through this node.
Case
KaTeX can only parse string typed expression
$f_{j} (L) < g (L)$ and $f_{j} (mid) < g (mid)$ :

This mirrors the first case with identical potential scenarios. However, since root-to-leaf traversal will definitely traverse this node for any query in the range
KaTeX can only parse string typed expression
$[L, mid)$ , the current node should be updated to $g (x) := f_{j} (x)$ to ensure the minimum value can be obtained.
Case
KaTeX can only parse string typed expression
$f_{j} (L) < g (L)$ and $f_{j} (mid) \geq g (mid)$ :

This implies that
KaTeX can only parse string typed expression
$p$ is located somewhere in $[L, mid)$ , and we need to recurse to the left child of the current node to handle that case. We don't need to update the current node as $g (x) \leq f_{j} (x)$ when $x \geq mid$ . Whenever a query of $mid \leq x < R$ is performed, the traversal from root to leaf is guaranteed to pass through this node.
Case
KaTeX can only parse string typed expression
$f_{j} (L) \geq g (L)$ and $f_{j} (mid) < g (mid)$ :

This implies that
KaTeX can only parse string typed expression
$p$ is located somewhere in $[L, mid)$ , and we need to recurse to the left child of the current node to handle that case. As the traversal from root to leaf is guaranteed to pass through this node whenever a query of $mid \leq x < R$ is performed, we need to update the current node $g (x) := f_{j} (x)$ so that we can retrieve the minimum value.

Below, the implementation for inserting the function is showcased. Bear in mind that the struct Node in this code is tailored to the problem described. Adjustments may be necessary based on specific needs.

using ll = long long;

struct Function {
  ll a, b, c, d;
  Function(ll a = 0, ll b = 0, ll c = 0, ll d = 0) : a(a), b(b), c(c), d(d) {}
  ll f(ll x) const { return a * x * x * x + b * x * x + c * x + d; }
};

const ll INF = 1e18;

struct Node {
  Function function{0, 0, 0, INF};
  Node *left = nullptr, *right = nullptr;
};

Node *root;
const int S = 350, L = S + 1, R = 1e5 + 1;

void insertFunction(Function newFunction, Node *node = root, int l = L,
                    int r = R) {
  if (!node)
    return;
  int m = (l + r) >> 1;
  bool left = newFunction.f(l) < node->function.f(l);
  bool mid = newFunction.f(m) < node->function.f(m);
  if (mid)
    swap(newFunction, node->function);
  if (l == r - 1)
    return;
  Node *&child = left != mid ? node->left : node->right;
  if (!child)
    child = new Node();
  insertFunction(newFunction, child, left != mid ? l : m, left != mid ? m : r);
}

Notice that the implementation is similar to Dynamic Segment Tree as it is using pointer and it allows a quite wide range interval of query. It is possible to deploy Li Chao Tree using the common array-based to store the node value. However, this comes at the expense of being restricted to a narrower index range.

Then, to determine which child node to traverse to, we simply follow the node that contains the range of the point of query. The implementation is as follows:

ll query(int x, Node *node = root, int l = L, int r = R) {
  if (!node || l > x || r <= x)
    return INF;
  int m = (l + r) >> 1;
  ll res = node->function.f(x);
  return x < m ? min(res, query(x, node->left, l, m))
               : min(res, query(x, node->right, m, r));
}

Solution

If we try to naively apply the polynomial function, we won't be able to meet the criteria of Li Chao Tree; consequently, this definitely results in Wrong Answer verdict. As melfice stated in his tutorial post, "a polynomial

KaTeX can only parse string typed expression

y = x^{3} + a x^{2} + b x + c

has at most one root greater than

k = max (∣ b ∣, ∣ c ∣) + 2

". Proof can be found here.

This implies that if we only consider

KaTeX can only parse string typed expression

[1 0^{5} + 2, 1 0^{5}]

(instead of

[0, 1 0^{5}]

) as our main range, then, we can effectively integrate the polynomial function into the Li Chao Tree. Then, for range

[0, 1 0^{5} + 2)

, we can simply update it by iterating it over as

1 0^{5} + 2 \approx 319

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  int T;
  cin >> T;
  while (T--) {
    int N, Q;
    cin >> N;
    vector<Function> functions(N);
    for (auto &f : functions)
      cin >> f.d >> f.c >> f.b >> f.a;
    cin >> Q;
    vector<int> t(Q);
    for (auto &x : t)
      cin >> x;

    root = new Node();
    vector<ll> ans(S + 1, INF);

    for (const auto &func : functions) {
      for (int x = 0; x <= S; ++x)
        ans[x] = min(ans[x], func.f(x));
      insertFunction(func);
    }

    for (const auto &x : t)
      cout << (x <= S ? ans[x] : query(x)) << "\n";
  }
  return 0;
}

Denote

KaTeX can only parse string typed expression

C

max (b_{m a x}, c_{m a x})

. Then, the overall time complexity for the implementation is:

O (N C + (N + Q) lo g (x_{m a x} - C))

My full solution is accessible here.

Li Chao Tree on Convex Hull Trick

While it is potent, the traditional Convex Hull Trick comes with a limitation: it requires the gradients to be strictly monotonous. One might adopt a dynamic approach by using

KaTeX can only parse string typed expression

std::set

instead of

std::vector

; however, this alternative introduces complexities, making it far from straightforward. Li Chao Tree is able to address those problems. If you are not familiar with the Convex Hull Trick yet, I suggest you to read my previous blog first. You can find it here.

In Convex Hull Trick, we may notice that for all pairs of functions, the number of intersection point is at most once. For affine functions

KaTeX can only parse string typed expression

f_{i} (x) = m_{i} x + c_{i}

and

f_{j} (x) = m_{j} x + c_{j}

(

1 \leq i < j \leq n

), the x-coordinate of the intersection point is located at

- \frac{m _{j} - m _{i}}{c _{j} - c _{i}}

. We may assign

p

with it and have the required condition of Li Chao Tree satisfied. Thus, the Convex Hull Trick can be implemented using the Li Chao Tree (although with a potential of having extra memory space). If you're curious about the practical application of this concept, you can check out my implementation for the problem discussed in the Convex Hull Trick post here.

References

melfice, "POLY - Editorial," CodeChef Discussion. https://discuss.codechef.com/t/poly-editorial/17335.
rama_pang, "[Tutorial] Li Chao Tree Extended," Codeforces. https://codeforces.com/blog/entry/86731.

← [Tutorial] DP Optimization: Convex Hull Trick [Tutorial] Lagrange Relaxation →

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