[Gym] XXI Open Cup. Grand Prix of Korea

4 года назад, # ^ |

+56

The problemset is designed for teams preparing ICPC Semifinals or World Finals.

→ Ответить

4 года назад, # |

Will there be a Div2 OpenCup contest?

→ Ответить

4 года назад, # ^ |

I sent it to Snark, but apparantly he didn't processed it. Div2 counterpart (KAIST 10th ICPC Mock Contest) will be uploaded to Gym shortly after.

→ Ответить

4 года назад, # ^ |

You mean shortly after the Div 1 starts today, or in a few days?

→ Ответить

4 года назад, # ^ |

← Rev. 2 →

I hope the former unless sth is broken.

I can't guarantee, so if you want team participation, it's better to schedule later.

→ Ответить

4 года назад, # ^ |

Thanks!

→ Ответить

4 года назад, # |

Is there Div 2 editorial?

→ Ответить

chemthan

4 года назад, # |

I: Subtract version. Only master is to calculate subtree sum quickly.

→ Ответить

Bohun

4 года назад, # |

+28

How to solve F?

→ Ответить

4 года назад, # ^ |

Interval graph is acyclic if there are at most 2 intervals at each point. If this is true then you can divide these alive intervals into two sets of nonoverlapping intervals. Hence you can solve this using maxflow mincost, where flow has value 2 and there are edges $$$i \to i+1$$$ with capacity $$$2$$$ and cost $$$0$$$ and $$$s \to e+1$$$ with capacity $$$1$$$ and cost $$$-w$$$ for each interval. To speed things up you need to compute potentials with dynamic programming instead of Bellman-Ford/SPFA and you can do this since this graph is acyclic.

→ Ответить

ecnerwala

4 года назад, # ^ |

← Rev. 5 →

Apparently you don't need to compute potentials quickly, we passed pretty comfortably with book SPFA.

EDIT For reference, we just used https://github.com/ecnerwala/cp-book/blob/40b6ce3f17a5b917531b44404fbc7d9e0d1e4cd7/src/mcmf.hpp, which is based on https://github.com/kth-competitive-programming/kactl/blob/3e39f71e55fb414540d518b11dc40216fe22fe1a/content/graph/MinCostMaxFlow.h but supports sparse graphs.

→ Ответить

4 года назад, # ^ |

We didn't :(

→ Ответить

4 года назад, # ^ |

← Rev. 2 →

I would say that your implementation is quite far from book SPFA, for me it's unintuitive that abusing priority queues in such a way is even possible :)

We tested all kinds of SPFA listed in Wikipedia, and any of them couldn't pass our data in <15s, which was in turn 100% random.

Also, I'm convinced that yours could provably run in $$$O((n + m) \log n)$$$ time given that the underlying graph is DAG.

→ Ответить

PavelKunyavskiy

4 года назад, # ^ |

+10

Doesn't this shortest path algorithm works exponentially long on test like

and so on?

I'm not sure if tests like this are constructible in min-cost-like problems, but seems to be strange. Anyway, it would probably be very fast on average.

→ Ответить

ecnerwala

4 года назад, # ^ |

I haven't looked in a while, but I thought it is upper bounded by O(VE). It might not be though...

→ Ответить

Benq

4 года назад, # ^ |

See subtasks 4 and 6 of https://tlx.toki.id/problems/apio-2013/C/

→ Ответить

ecnerwala

4 года назад, # ^ |

← Rev. 2 →

Ah, I remember how it works. In SSPA, if all edges are initially positive cost, then we maintain a potential function which ensures they stay positive, so the runtime is just Dijkstra. If the edge costs can be initially negative, you should run a first pass of Bellman Ford to initialize the potential function (which I seem to be missing, but KACTL has).

Our implementation adjusted the edge costs to implicitly choose a potential function of -x*10**12, just like Petr suggested.

→ Ответить

PavelKunyavskiy

4 года назад, # ^ |

If all edges are initially positive, it's okey. I thought you are claiming your code to work with negative edges too, and was a bit surprised.

→ Ответить

Petr

4 года назад, # ^ |

+60

We just set the initial potentials to -x*10**12, no need for dynamic programming :)

→ Ответить

4 года назад, # |

+57

Thank you for your participation!

→ Ответить

Golovanov399

4 года назад, # |

How to A properly? Our solution is:

For two vectors a and b the maximal number of things we can present is max flow of the complete bipartite graph with unit edges between the sides, and with edges from source to left with capacities $$$a_i$$$ and from right to sink with capacities $$$b_i$$$. Or, equivalently, min cut. If we take $$$k$$$ vertices from the left side into the mincut and $$$m - l$$$ vertices from the right side, then we need to check if the sum of maximal $$$k$$$ $$$a_i$$$-s plus sum of maximal $$$(b_j - k)$$$-s minus sum of all $$$b$$$-s can be greater than zero, or something. This is a segment tree with += on subseg and getmax on the whole tree. Did 46 teams implement the exact this? I think it's quite tricky to avoid mistakes here.

→ Ответить

4 года назад, # ^ |

I implemented that. It's not mandatory to interpret such as maxflow, but I don't know any other solution that has a different implementation.

→ Ответить

ppavic

4 года назад, # ^ |

Our interpretation was something similar to the Erdos-Gallai theorem so the implementation wasn't as tricky, but we had a lot of failed submissions because of other reasons xD

→ Ответить

4 года назад, # ^ |

Quite standard problem and approach I would say. Moreover you can copy and paste this segment tree to E and L!

→ Ответить

Radewoosh

4 года назад, # ^ |

Yea, today I've used the same segment tree in 3 problems xd

→ Ответить

YangDavid

4 года назад, # ^ |

+10

This problem is almost the same as the bipartite graph realization problem. Our team googled and found out the Gale–Ryser theorem. It's not hard to prove that after changing the condition of $$$\sum\limits_{i=1}^n a_i = \sum\limits_{i=1}^n b_i$$$ to $$$\sum\limits_{i=1}^n a_i \leq \sum\limits_{i=1}^m b_i$$$, the theorem works for this problem. The remaining part is just using a segment tree to keep track of the inequalities.

→ Ответить

SolveMachine

4 года назад, # |

Where to find problems?

→ Ответить

Fivee

4 года назад, # |

How could one upsolve div2 contest ?

→ Ответить

4 года назад, # |

E was nice, but almost identical to this problem: https://mirror.codeforces.com/contest/1109/problem/F

→ Ответить

4 года назад, # ^ |

I even have a TLE attempt on this problem, what a shame. Sorry!

Actually, the problem had an underlying graph as a tree, but we have changed it after I remembered a problem in Ptz which was exactly the same. Though, the model solution was a complicated $$$O(n \log^2 n)$$$ one, and my solution was n^2.

→ Ответить

yosupo

4 года назад, # |

+35

I love this set, thanks!

→ Ответить

Golovanov399

4 года назад, # |

← Rev. 2 →

+40

Our solution in E was in $$$O(n\log{n})$$$ time, yet no link-cut or other fancy structures were required; the main part of our solution was to find such $$$r$$$ for each $$$l$$$ that the induced subgraph on vertices $$$[l, r]$$$ is a set of isolated paths and vertices; this is done by maintaining the paths as a set of treaps (each path is an in-order traversal of some treap) and then moving two pointers, after this some segment tree.

→ Ответить

gamegame

4 года назад, # ^ |

← Rev. 5 →

+18

300iq provided a fix to my idea. So we can use the Queue Undo Trick for this idea, by imagining that we insert and pop vertices, as here inserting vertices is $$$O(1)$$$ updates to dsu, but we cannot use this to replace the LCT in editorial.

→ Ответить

4 года назад, # |

+61

The contest is ready in Codeforces Gym. Enjoy!

→ Ответить

lin7xu

2 года назад, # |

← Rev. 2 →

+15

A set of interval covers each point at most K times if and only if it can be partitioned into K disjoint set of intervals. You can prove this by greedy algorithm or using the fact that interval graphs are perfect.

What does perfect mean in the editorial?

→ Ответить

lin7xu

2 года назад, # ^ |

← Rev. 2 →

This can directly proved by Dilworth theorem

→ Ответить