# LineContainer

Authors: Benjamin Qi, Andi Qu

Convex Containers

## Half-Plane Intersection

Resources | ||||
---|---|---|---|---|

CF | ||||

Petr | Expected linear time! |

### This section is not complete.

Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|

Kattis | Normal | ||||

JOI | Hard | ||||

Balkan OI | Very Hard | ## Show TagsBinary Search, Geometry |

## $\mathcal{O}(N \log N)$ CHT)

LineContainer (akaStatus | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|

YS | Normal |

Resources | ||||
---|---|---|---|---|

KACTL | source of code that I (Ben) use | |||

cp-algo | related topic (but not the same) |

### Example Problem

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

#### Analysis

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 $dp[i]$ be the minimum cost to build the bridge so that the last build pillar is pillar $i$.

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

Notice how

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

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

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

Code

## Problems

Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|

YS | Normal | ||||

POI | Normal | ## Show TagsConvex, DP | |||

CEOI | Hard | ## Show TagsConvex, DP | |||

FHC | Hard | ## Show TagsConvex, DP | |||

AC | Hard | ||||

TLX | Hard | ||||

Old Gold | Hard |

### Module Progress:

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