Authors: Benjamin Qi, Andi Qu

Convex Containers

Half-Plane Intersection

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Example problem + implementation?
StatusSourceProblem NameDifficultyTags
Balkan OIVery Hard
Show TagsBinary Search, Geometry

LineContainer (aka O(NlogN)\mathcal{O}(N \log N) CHT)

StatusSourceProblem NameDifficultyTags

source of code that I (Ben) use


related topic (but not the same)

Example Problem

Focus Problem – try your best to solve this problem before continuing!


Instead of focusing on the pillars that should be destroyed, let's instead focus on the pillars that remain.

The total cost consists of the cost due to height differences plus the cost of destroying unused pillars. The latter cost is equal to the cost to destroy all pillars minus the cost to destroy the remaining pillars.

Since the cost to destroy all pillars is constant, we can thus turn the problem into one about building pillars instead of destroying them!

From this, we get a basic DP recurrence. Let dp[i]dp[i] be the minimum cost to build the bridge so that the last build pillar is pillar ii.

dp[1]=w1dp[1] = -w_1 and the following recurrence holds:

dp[i]=minj<i(dp[j]+(hjhi)2wi)=minj<i(dp[j]+hj22hihj)+hi2wi\begin{aligned} dp[i] &= \min_{j < i}(dp[j] + (h_j - h_i)^2 - w_i)\\ &= \min_{j < i}(dp[j] + h_j^2 - 2h_ih_j) + h_i^2 - w_i \end{aligned}

Notice how

dp[j]+hj22hihjdp[j] + h_j^2 - 2h_ih_j

effectively describes a linear function y=mx+cy = mx + c, where m=2hjm = -2h_j, x=hix = h_i, and c=dp[j]+hj2c = dp[j] + h_j^2

This means that we can use CHT to compute dp[i]dp[i] efficiently!

However, since mm is not monotonic, we can't use linear CHT using a deque, so we must settle with O(NlogN)\mathcal{O}(N \log N).

I implemented CHT using a std::set here, but other implementations using things like the Li-Chao tree should work similarly.

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
struct Line {
bool type;
double x;
ll m, c;


StatusSourceProblem NameDifficultyTags
Show TagsConvex, DP
Show TagsConvex, DP
Show TagsConvex, DP
Old GoldHard

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