Google Code Jam Round 1B

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Блог пользователя kingofnumbers

Google Code Jam Round 1B

Автор kingofnumbers, история, 9 лет назад, По-английски

Hi,

This is to remind you that Google Code Jam Round 1B will be tomorrow at this time, Top 1000 will advance to 2nd round.

let's discuss the problems after the contest.

Good luck.

kingofnumbers
9 лет назад
52

Комментарии (49)

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king_of_math

9 лет назад, # |

← Rev. 2 →

How to register the contest?

Edit: "Coding has commenced! Registration is now closed." <== Okay :/

→ Ответить

TimonKnigge

9 лет назад, # ^ |

You can participate if you qualified during the Qualification Round.

→ Ответить

aka.Sohieb

9 лет назад, # |

+26

starts in 10 mins.
GL & HF

→ Ответить

anh1l1ator

9 лет назад, # |

-7

Very Bad round for me . I didn't understand why my solution fails on B . B had 40+ % accuracy. I tried to be Greedy it failed( At the back of my mind I had a kind of proof).

Then I did a simply DP on integers but yet I recieved many WAs. Code .

Can someone provide some trivial test ? I think my understanding of the statement is wrong!

→ Ответить

Beresta

9 лет назад, # ^ |

Well I solved in on small test, but failed on big one. Not sure why — maybe my solution is wrong, or just implementation (it is kinda complex, easy to make a mistake somewhere).

→ Ответить

dush1729

9 лет назад, # ^ |

Small passed. I did it by brute force.

→ Ответить

shubham1694

9 лет назад, # ^ |

Input:

?6 31

Expected Output:

26 31

→ Ответить

Ishan.nitj

9 лет назад, # ^ |

-8

How to solve second question for large input.

→ Ответить

hellman_

9 лет назад, # ^ |

+16

Go from left to right and try to assign digits either equal, or differing at most by 1 (to minimize the distance). If at some point prefixes differ, then you have to assign the rest free digits to 9 for the string with smaller prefix and assign all free digits to 0 for the other string.

Otherwise, if you assign equal digits or they are given equal, just recurse further.

Also, if you assign equal digits when both strings have "?" in this position, check only 0 0, it will minimize both scores.

In such way the recursion never branches (or you can think that it has branches of length 1).

→ Ответить

brighterstill

9 лет назад, # ^ |

1 ?5 10 This was in the test cases for small.

→ Ответить

dush1729

9 лет назад, # |

How to do C small?

→ Ответить

Beresta

9 лет назад, # ^ |

← Rev. 2 →

You can bruteforce it with simple recursion (trying to pick all possible combinations of valid participants). With careful implementation it passes.

→ Ответить

rajat1603

9 лет назад, # ^ |

for Csmall write a bitmask DP by storing a mask of topics already used. Complexity O(2^N * N * N).

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Fcdkbear

9 лет назад, # ^ |

You can use dynamic programming. The state is the set of all already placed topics (represented as binary mask). To make a transition, just iterate through all possibilities to place a new topic, and check if it is faked. The complexity is O(2ⁿ * n²)

→ Ответить

Maestr0

9 лет назад, # ^ |

Split topics into 2 sets — fake and real. Then for each fake topic check if there are first and second word in 'real' topics set. Try all possible splits, 2^N for each test.

→ Ответить

xlwinner11

9 лет назад, # ^ |

It's simple: just check all the 2^n masks, O(n^2) for each mask.

How to check it? For each topic marked as "fake" in this mask you should check whether the first word appears in the list of unfaked topics. Same for the second word. If both words do not appear — this mask can't be possible answer.

→ Ответить

james007

9 лет назад, # ^ |

My brute-force solution with complexity O(2ⁿ * n) following:

Try all possible 2ⁿ mask, where 1 means, that topic if fake.
The answer is maximum of bitcount(mask) where for each fake topic there are not fake topics with same first and second words. This check procedure can be done with O(n) complexity.

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Ra1nWarden

9 лет назад, # ^ |

I did it with minimum edge cover. Build a bipartite graph and each entry is an edge connecting two vertices from the two sets. The answer is the total number of edges minus the minimum edge cover.

→ Ответить

NegaTeeF

9 лет назад, # |

+16

Analysis can be found here.

→ Ответить

eduardische

9 лет назад, # ^ |

← Rev. 2 →

+58

I have a different solution to C than the analysis, and it does not involve being greedy at any stage. Consider the example they gave:

HYDROCARBON COMBUSTION
BIOMASS COMBUSTION
QUAIL COMBUSTION
QUAIL BEHAVIOR
QUAIL CONTAMINATION
GROUNDWATER CONTAMINATION
GROUNDWATER HYDROLOGY

Now, instead of finding minimum edge cover on the top graph, I simply find the maximum flow in the network below.

The idea behind this is as follows. Let a single unit of flow represent a fake topic. Then it has to go through the middle edge, and only a single unit of flow can pass there. However, we have the additional restriction. Every word has to be in at least one legit topic. To enforce that the capacity of edges representing the words are their frequency minus one.

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saurabhverma

9 лет назад, # ^ |

Can you show how you have enforced last condition by subtracting capacity by one? What is the final answer if flow is F. Isn't it F?

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Chortos-2

9 лет назад, # ^ |

The topics contained in the flow found here are fakes, while the remaining topics are real.

By reducing the capacity of each word by one, you ensure that each word has at least one unused topic remaining after the fake flow is constructed, i. e. that each word has at least one real topic.

Imagine that after finding the fake flow, you subtract the fake flow from the graph, remove the word capacity bounds (but keep the topic capacity bounds) and find the maximum flow in the resulting graph. Then you’ll get the real flow, and it will incorporate all remaining topics due to the unbounded word capacities. By ensuring that this graph contains at least one topic involving each word, you ensure that the real topics cover all words, which is good. Conversely, if you hadn’t reduced the word capacities by one before finding the fake flow, the fake flow might have taken up all topics involving a particular word and you’d be left with no real topic covering that word, which would be bad.

Of course, you want to minimize the number of real topics, so you want to minimize the number of edges remaining after subtracting the fake flow (because each remaining edge corresponds to a real topic), so you want to maximize the number of edges in the fake flow, so you want the fake flow to be maximal.

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Al.Cash

9 лет назад, # ^ |

+10

I don't understand why they mentioned greediness. The answer is simply n — (number of different first words) — (number of different second words) + (size of maximum matching).

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TimonKnigge

9 лет назад, # ^ |

I'm guessing it's because rather than explaining how to get the size of the edge cover, they also wanted to explain how to construct the edge cover itself.

→ Ответить

kingofnumbers

9 лет назад, # |

+59

when I see those large number of participants in GCJ I wonder are there a lot of competitive programmers who don't participate neither in CF nor in TC but only in GCJ?

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I_love_Tanya_Romanova

9 лет назад, # ^ |

+28

I know a lot of guys who were active at CF/TC/CodeChef in past, and nowadays only participate in annual competitions like FHC, TCO, GCJ etc. A lot of people are active at online judges while preparing for ICPC, and then only participating in major competitions later.

And also if you are considering active contestants, still a lot of them are only attending some small percentage of CF/TC contests, while trying not to miss events like GCJ.

→ Ответить

hellman_

9 лет назад, # ^ |

+22

Also GCJ is quite friendly for languages like Python or even Sage, since you run everything locally and the time limit is a few minutes.

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kingofnumbers

9 лет назад, # ^ |

+19

but this reasoning is still not enough to cover the very large numbers of participants in GCJ, if you look on the number of participants of qualification round it is around 27,000 and 22,000 out of them qualified to round 1, which is even more than total number of accounts in CF without even removing the fake accounts :D

also looking at the difficulty of today's round and the number of points required to qualify to 2nd round I can say that one need to be div1 coder in CF to be able to qualify to 2nd round (or have equivalent skills). 3,000 coders will advance to 2nd round which is more than the number of div1 accounts in CF, again without removing fake accounts :D

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Edvard

9 лет назад, # ^ |

← Rev. 2 →

+13

I think one of the main reasons is Googles brand (second place in Forbes rating after Apple). Every programming contest is positioning as a place for hiring of coders and about every coder wants to work at Google (of course the brand is not the only reason for that). So in GCJ are participating a lot of users who are not competing in other contests at all.

→ Ответить

antonkov

9 лет назад, # ^ |

Google become most valuable company in the world not long time ago http://qz.com/607396/google-has-officially-dethroned-apple-as-the-worlds-most-valuable-company/

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Edvard

9 лет назад, # ^ |

Actually I'm talking only about the Google brand. It seems I was wrong and Google on the third place according to Forbes rating.

→ Ответить

Ishan.nitj

9 лет назад, # |

-21

HOW to solve the 2nd question for large input.

→ Ответить

Edvard

9 лет назад, # |

+45

Very nice problems, as usual on GCJ.

→ Ответить

MPeti

9 лет назад, # |

Could someone upload a correct solution for the large practice data for task B? I've implemented mostly the same thing the analysis said, and the output does seem correct for the cases I can easily check by hand, but it's apparently incorrect for some. Probably a corner case...

→ Ответить

Edvard

9 лет назад, # ^ |

+13

You can download a solution of any participant in the scoreboard page: just set Solution Download checkbox.

→ Ответить

j11porcelli20

9 лет назад, # |

-34

vi find(int freqs[]){
	vi r;
	while(freqs['Z'-'A']-1 >= 0 && freqs['E'-'A']-1 >= 0 && freqs['R'-'A']-1 >= 0 && freqs['O'-'A']-1 >= 0){
		freqs['Z'-'A']--; freqs['E'-'A']--; freqs['R'-'A']--; freqs['O'-'A']--;
		r.PB(0);
	}
	while(freqs['O'-'A']-1 >= 0 && freqs['N'-'A']-1 >= 0 && freqs['E'-'A']-1 >= 0){
		freqs['O'-'A']--; freqs['N'-'A']--; freqs['E'-'A']--;
		r.PB(1);
	}
	while(freqs['T'-'A']-1 >= 0 && freqs['W'-'A']-1 >= 0 && freqs['O'-'A']-1 >= 0){
		freqs['T'-'A']--; freqs['W'-'A']--; freqs['O'-'A']--;
		r.PB(2);
	}
	while(freqs['T'-'A']-1 >= 0 && freqs['H'-'A']-1 >= 0 && freqs['R'-'A']-1 >= 0 && freqs['E'-'A']-2 >= 0){
		freqs['T'-'A']--; freqs['H'-'A']--; freqs['R'-'A']--; freqs['E'-'A']-=2;
		r.PB(3);
	}
	while(freqs['F'-'A']-1 >= 0 && freqs['O'-'A']-1 >= 0 && freqs['U'-'A']-1 >= 0 && freqs['R'-'A']-1 >= 0){
		freqs['F'-'A']--; freqs['O'-'A']--; freqs['U'-'A']--; freqs['R'-'A']--;
		r.PB(4);
	}
	while(freqs['F'-'A']-1 >= 0 && freqs['I'-'A']-1 >= 0 && freqs['V'-'A']-1 >= 0 && freqs['E'-'A']-1 >= 0){
		freqs['F'-'A']--; freqs['I'-'A']--; freqs['V'-'A']--; freqs['E'-'A']--;
		r.PB(5);
	}
	while(freqs['S'-'A']-1 >= 0 && freqs['I'-'A']-1 >= 0 && freqs['X'-'A']-1 >= 0){
		freqs['S'-'A']--; freqs['I'-'A']--; freqs['X'-'A']--;
		r.PB(6);
	}
	while(freqs['S'-'A']-1 >= 0 && freqs['E'-'A']-2 >= 0 && freqs['V'-'A']-1 >= 0  && freqs['N'-'A']-1 >= 0){
		freqs['S'-'A']--; freqs['E'-'A']-=2; freqs['V'-'A']--; freqs['N'-'A']--;
		r.PB(7);
	}
	while(freqs['E'-'A']-1 >= 0 && freqs['I'-'A']-1 >= 0 && freqs['G'-'A']-1 >= 0 && freqs['H'-'A']-1 >= 0 && freqs['T'-'A']-1 >= 0){
		freqs['E'-'A']--; freqs['I'-'A']--; freqs['G'-'A']--; freqs['H'-'A']--; freqs['T'-'A']--;
		r.PB(8);
	}
	while(freqs['N'-'A']-2 >= 0 && freqs['I'-'A']-1 >= 0 && freqs['E'-'A']-1 >= 0){
		freqs['N'-'A']-=2; freqs['I'-'A']--; freqs['E'-'A']--;
		r.PB(9);
	}
	return r;
}

Can somebody tell me what is wrong with this logic for problem A? (Besides the fact it is quite cumbersome), thanks

→ Ответить

j11porcelli20

9 лет назад, # ^ |

-8

Also I have looked at a few solutions but I still dont get the idea those people have taken

→ Ответить

MPeti

9 лет назад, # ^ |

Have you read the problem analysis? https://code.google.com/codejam/contest/11254486/dashboard#s=a&a=0

→ Ответить

MPeti

9 лет назад, # ^ |

Input string: FOURNINE Your program will take out a ONE, leaving FURNI, which can't be decomposed further.

→ Ответить

eduardische

9 лет назад, # ^ |

As analysis say, you need to ensure that the order of your greedy is correct, which is not in this case. In the example TWOSEVEN, there are enough letters to take ONE and screw you over.

→ Ответить

Swistakk

9 лет назад, # |

+210

Who invented that stupid upper bound on number of friends which is 30 -_-? Do Google's employees have less than 30 friends?

→ Ответить

chuka231

9 лет назад, # ^ |

+75

that is why google could not compete with facebook.

→ Ответить

Zlobober

9 лет назад, # ^ |

+108

This is a way for you to decide who are you your real buddies.

→ Ответить

Fear_Is_An_Illusion

9 лет назад, # ^ |

+21

Actually one page fits 30 people, so that's why.. :P

→ Ответить

Gassa

9 лет назад, # ^ |

+22

Not actually correct. Back when one page of the full scoreboard contained only 25 entries, the number of friends was already 30.

→ Ответить

Xellos

9 лет назад, # ^ |

I also think it's a stupid restriction, but since I can't keep track of even 30 people's performances anyway, I don't mind all that much.

→ Ответить

MPeti

9 лет назад, # |

Having qualified in this round, is it still going to be possible to take part in round 1C out of competition in real time? Or will I be limited to submitting after it's over?

→ Ответить

ivanilos

9 лет назад, # ^ |

+17

Limited to submit only when it is over if you have passed 1A or 1B

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WasylF

9 лет назад, # |

Скажите, пожалуйста, фраза "Emails with official results will go out early next week." переводится на русский как: "официальных результатов вы не дождетесь"?

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