Google Code Jam 2018 Round 2

7 лет назад, # ^ |

+14

Google doesn't care about CF (or particularly about competitive programming in general), you should write to the Codejam Google Group. Although I don't know how much requests there are watched...

One guy had broken round 1C, the dashboard ignored timezones and had the round "start" 2 hours earlier — the problems weren't visible for those 2 hours, so he had 30 minutes to compete. This bug only happened at one computer, but it was pretty persistent (restarting browser or cache clean didn't work).

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hex539

7 лет назад, # ^ |

← Rev. 2 →

+34

It's more efficient to email codejam@google.com if you found an issue, especially if it's time-sensitive.

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we7d

7 лет назад, # ^ |

← Rev. 2 →

+24

they don't want to fix it, I emailed them about the issue last round as my contest ended after 1:30 instead of 2:30.

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7 лет назад, # |

← Rev. 3 →

+50

Had a lot of ideas for D, cant think of how to keep my code clean and effective .... Should've gone for small instead of going for large + AK. :/

Edit:

Me and my teammates after the contest:

Me: Whew, this is the first time I ever used the simplex template, it actually fit into B's TL!

Teammate: Isn't it DP?

Me: What?

Teammate: What?

Edit 2: So it turns out I am the only edgy kid who used simplex for B ... ?

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kwangg

7 лет назад, # ^ |

+17

How did you use simplex for B?

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https://ideone.com/VA2Uvw

7 лет назад, # ^ |

Maximize dot(c, x):

c is a column vector filled with 1, each entry of x represents if we will pick (x reds, y blues), so dot(c, x) will be the amount of jugglers we have at the end.

Simliar to the editorial make sure you only keep pairs that are possible to be juggled with the amount of chainsaws we have.

While satisfying Ax <= b:

First two rows of A will be representing the constraints on how much red / blue chainsaws we have, if the i-th entry of x represents (x reds, y blues), then A[0][i] = x, A[1][i] = y. Set b[0] = Red, b[1] = Blue, such that we will never exceed the budget.

For the i-th entry in x, to make sure we don't have jugglers which share the same juggling arrangement, add extra rows where A[i+2][i] = b[i+2] = 1.

Code:

Simplex template credit goes to MIT 2008 ACM template.

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7 лет назад, # ^ |

Can you please elaborate a little on last part to ensure that we don't have jugglers sharing same arrangement?

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7 лет назад, # ^ |

It makes sure max(x) <= 1, similar to how you prove IP is NPC from knapsack.

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7 лет назад, # ^ |

Makes sense now. Thank you :)

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hellman_

7 лет назад, # ^ |

+34

How can you be sure that the answers from simplex solver are integers?

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7 лет назад, # ^ |

← Rev. 2 →

I am not sure about the proof -- But I suppose that the problem nature encourages the simplex solver to take pairs which spends less chainsaw to attain maximimzed real results, and taking floor shouldn't hurt.

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7 лет назад, # |

+32

239 people with 100 points... that's a lot more than last year's 7.

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mahmud2690

7 лет назад, # |

Was C Augmenting Path/Max-Flow?

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rajat1603

7 лет назад, # ^ |

just bipartite matching(n^2 — match matching for each value)

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mahmud2690

7 лет назад, # ^ |

-14

OMG, it seems I copied not-working code from the internet).

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7 лет назад, # |

← Rev. 2 →

For C, why the greedy idea (Let s the sum of size of maximum matching for each number, then answer = n² - s) fail?

I regret wasting too much time on C, not realizing that D is actually easier than C.

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zscoder

7 лет назад, # ^ |

That's the correct answer for C.

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7 лет назад, # ^ |

+40

Then what the ... (trying my best not to speak bad language) is wrong with my solution for C

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Rafbill

7 лет назад, # ^ |

← Rev. 3 →

+116

You forget to reset visited between different DFS calls.

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7 лет назад, # ^ |

+320

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7 лет назад, # ^ |

Except it should not fail... I suppose you built a flow graph w.r.t row/column for each number?

Meanwhile me failing to code D.

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7 лет назад, # ^ |

+103

IMO C was the easiest problem today :P

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Andrei1998

7 лет назад, # ^ |

Finally, someone who shares my opinion.

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7 лет назад, # ^ |

Could you explain your solution?

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7 лет назад, # ^ |

+17

Iterate x from - n to n. For each x, we will find the maximum number of cell with number x that we can keep unchanged. To do that, build a bipartite graph, each vertex on the left right represent a row, each vertex on the right side represent a column. Add an edge from u on the left to v on the right if a[u][v] = x. Then, the number of cell we can keep is the cardinality of maximum matching in the above graph.

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7 лет назад, # ^ |

← Rev. 2 →

I just don't why finding a maximum matching in above graph correspond to maximal independent set in original graph.

What's the meaning of taking edge ij?

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7 лет назад, # ^ |

← Rev. 2 →

Each edge is corresponding to a cell in the original graph. In the maximum matching, no two edge coincident to the same vertex, similar to how no two chosen cells should be in the same row or column.

Consider the following example

0 3 0 
0 3 0
0 3 3

If we consider number 3, then the bipartite graph will look like the following:

1  1
 \ 
  \
2--2
  /
 /
3--3

One possible maximum matching is (1, 2), (2, 2) and (3, 3), which corresponding to the solution of keeping cell (1, 2), (2, 2) and (3, 3) unchanged.

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https://codejam.herokuapp.com/

7 лет назад, # ^ |

Thank you so much.

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---Grigor---

7 лет назад, # |

← Rev. 2 →

+12

Is there any scrapper available online to filter contest results by countries at least?

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kalimm

7 лет назад, # ^ |

+22

It's currently not available for this contest but some guy from CF updates this site regulary.

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Jakube

7 лет назад, # ^ |

+37

Here is one: http://gcj.aditsu.net/scores?r=2018-2

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galloska

7 лет назад, # ^ |

There is also this one created by Diego1149 :)

https://goo.gl/1uwya8

More explanation here: http://mirror.codeforces.com/blog/entry/58941

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7 лет назад, # |

How to solve C? I thought that we should compute minimum vertex cover in the graph where we have edges between vertex with the same color and (same row or column) but this graph is not bipartite. :(

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anh1l1ator

7 лет назад, # |

B large Fail Test Case?

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cdkrot

7 лет назад, # |

+49

Wow, compilation error still counts as penalty attempt.

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7 лет назад, # ^ |

Yeah, I also got one in previous rounds for using C++17.

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steiner

7 лет назад, # ^ |

I believe my OK large test submission was obscured by CE submissions after it. Is it really ok?)

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Nirma_Sucks

7 лет назад, # |

Can someone give me a hint why this code is failing in A : https://ideone.com/hVQGi2

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eulerscheZahl

7 лет назад, # ^ |

+14

At this point I like to generate random input to see if it crashes.

1
6
3 2 0 0 0 1
Exception in thread "Main" java.lang.ArrayIndexOutOfBoundsException: -1
	at Solution.run(Solution.java:207)
	at java.lang.Thread.run(Thread.java:748)

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Nirma_Sucks

7 лет назад, # ^ |

Ty very much:)

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T1duS

7 лет назад, # |

How to solve D?

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7 лет назад, # ^ |

I think this should work:

Notice that if you go deep enough, any fixed-size pattern can only be part of a very simple grid composed of 4 quadrants with the same colours (let's allow a quadrant to have colour None to deal with edge cases). Each such grid is formed by expanding a square 2x2 in the original grid sufficiently many times; there are at most 81 of them, so we can find all such colourings by brute force.

Now, we can try dividing the original grid into 4 parts by a vertical and horizontal line; there are O(RC) ways to do this, we can try them all. We can also try all possible colourings. For each of these cases, we know which cells can't be in our pattern, and out of the remaining cells, we can build the largest connected component. Then we just need the maximum of all CC. Time O(R²C²).

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KarlisS

7 лет назад, # ^ |

Using 2⁴ = 16 coloring isn't too difficult either. You just have to expand 1x1 and 1x2 colorings to matching 2x2 coloring when checking which ones are present.

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7 лет назад, # ^ |

In the end, my implementation could just ignore colourings with None, so not even expanding anything was necessary, but using dummy objects that simplify implementation and only increase the constant factor is a good approach in many problems.

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hellman_

7 лет назад, # ^ |

-8

My python implementation TL'd on large, so sad. 100 tests x 16 colourings x R^2C^2 = 256 x 10^6. Didn't have time to generate maxtests, but was pretty sure that should fit...

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7 лет назад, # |

← Rev. 3 →

+85

Has it happened to you that a solution thats calculates all the 500² answers and then solves each test in constant time passed the small, but not the large? What is wrong with their platform? How is that supposed to work?

Code:

Spoiler

#include <bits/stdc++.h>

using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    vector<vector<int>> dp(501, vector<int>(501, 0));
    for (int i = 0; i <= 500; ++i) {
        for (int j = 0; j <= 500; ++j) {
            if (i == 0 && j == 0) continue;
            for (int k = 500 - i; k >= 0; --k) {
                for (int l = 500 - j; l >= 0; --l) {
                    dp[i + k][j + l] = max(dp[i + k][j + l], dp[k][l] + 1);
                }
            }
        }
    }

    int t; cin >> t;
    for (int tt = 1; tt <= t; ++tt) {
        int a, b; cin >> a >> b;
        cout << "Case #" << tt << ": " << dp[a][b] << '\n';
    }

    return 0;
}

Verdict is: AC on small, TLE on large.

Locally runs on just over 9s.

Their TL is 25s.

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7 лет назад, # ^ |

There are probably many reasons why your code can get large wrong while passing small. What you have described here is definitely too less to draw any conclusions.

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7 лет назад, # ^ |

Fair enough. I've added some information to my comment.

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7 лет назад, # ^ |

OK, if that is TLE then it is definitely weird :P. Hypothetical WA could have been easily your fault.

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7 лет назад, # ^ |

+22

Btw, I know that is irrelevant, but you can change

for (int i = 0; i <= 500; ++i) {
        for (int j = 0; j <= 500; ++j) {

for (int i = 0; i <= 33; ++i) {
        for (int j = 0; j <= 33; ++j) {

and your code should remain correct.

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7 лет назад, # ^ |

← Rev. 2 →

+51

I know, I figured it out later during the contest, but I did not resubmit, due to obvious reasons XD.

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7 лет назад, # ^ |

+61

Yesterday, I'm pretty sure I solved a problem by resubmitting (1B B, I was trying an alternative solution that was getting TLE sometimes). And in 2B here, I got TLE on something that was running in 4 seconds locally, even though the time limit on the server is much larger.

Maybe they're measuring time with an hourglass or something?

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7 лет назад, # ^ |

+14

Well, that's definitely an unpleasant experience to have during actual contest. Maybe they should pay more attention to testing their platform more thoroughly before throwing it to the world.

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7 лет назад, # ^ |

← Rev. 2 →

+133

Update: Resubmitting after the contest ended, the verdict was AC on small and large. Guess you have to submit at the right moment to do well on this contest.

How stupid.

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7 лет назад, # ^ |

+121

Try making some noise about it to organizers. They should be aware of platform's weaknesses. It's not only your business, we too don't want such surprises in future :D.

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7 лет назад, # ^ |

+34

I have written an e-mail to them. I hope they will catch up with the issue, and it won't happen in the future. Thanks for the idea.

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kwangg

7 лет назад, # ^ |

+31

Damn, O(n^4) got AC? Nice.

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Asgar

7 лет назад, # ^ |

For me, 501x501 array solution first gave MLE :) on the small test set, then exact the same code passed smaller, then gave MLE on big test set, then third time, gave WA on big test set. But WA was correct.

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grekiki

7 лет назад, # |

Any "hacks" for falling balls? Had wrong answer but haven't figured it out yet.

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shadowfax

7 лет назад, # ^ |

If you are seeking tests for the small:

1
5
2 0 0 1 2

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galloska

7 лет назад, # ^ |

← Rev. 3 →

Try this one...

1
7
1 0 0 5 0 0 1

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7 лет назад, # |

← Rev. 2 →

I can't understand official solution for c. What's the intuition behind taking at most one vertex from edge AiBj?

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cr0n

7 лет назад, # |

← Rev. 2 →

Getting wrong answer for B, but then I took someone else's code that was accepted, generated cases for all possible R & B and took a diff of the output with my code. Surprisingly it's the same. Edit: Figured it out, there was an overflow while calculating the dp

Spoiler

#include <bits/stdc++.h>
using namespace std;
#define LIMIT 512

int dp[LIMIT][LIMIT][LIMIT];
bool vis[LIMIT][LIMIT][LIMIT];
int build(int x,int y,int z){
	if(vis[x][y][z]) return dp[x][y][z];
	vis[x][y][z] = true;
	int maxval = 0;
	for(int i=1,j=0;y-x*i >=0 && z-j>=0;j+=i,i++) maxval  = max(maxval,i+build(x+1,y-x*i,z-j));
	//cout<<x<<" "<<y<<" "<<z<<" "<<maxval<<endl;
	return dp[x][y][z] = maxval;
}

int main(){
	for(int i=0;i<LIMIT;i++) for(int j=0;j<LIMIT;j++) for(int k=0;k<LIMIT;k++) build(i,j,k);
	int cases;cin>>cases;
	for(int n=0;n<cases;n++){
		int r,b;cin>>r>>b;
		cout<<"Case #"<<n+1<<": "<<build(0,r,b)-1<<endl;
	}
	return 0;
}

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chemthan

7 лет назад, # |

← Rev. 2 →

Now, The tests are assumed that can be all maximum? In the past, there are small testcases and a few large testcases.

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https://docs.google.com/spreadsheets/d/10q6bxqgnd0U9KNzGFvAl7Ik6kWZ8_wbfiUMRN-cJdXE/edit?usp=sharing

7 лет назад, # ^ |

In some problems in rounds 1, the constraints were in the format "3 out of 100 tests are maxtests, the rest are small", so you should probably assume all tests are equally large unless written otherwise. It makes sense if you need to pass test i to pass test i + 1.

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varunvora

7 лет назад, # |

+19

Scraped scoreboard for all rounds so far in GCJ 2018.

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Jatana

7 лет назад, # |

+47

For fucks sake, I didn't solve task because forget to put '#' in 'Case #x:'
I got WA and didn't understand why.
So, suicide is the only way.

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Jakube

7 лет назад, # ^ |

+59

Another way: Copy the expected output and diff it with your output. (I also learned that the hard way.)

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7 лет назад, # ^ |

Calm down dude

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mickeyandkaka

7 лет назад, # ^ |

same with you with 6 wrong submissions for problem A

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7 лет назад, # |

For "Graceful Chainsaw Jugglers" problem, I was thinking of a greedy solution. It is as follows:

Make all possible pairs such {A,B} s.t. A+B = i. Here we iterate i from 1 to 1000, and if (R-A_k) ≥ 0 and (B-B_k) ≥ 0, then we update ans = (ans + 1), R = (R - A_k) and B = (B - B_k).

I got a WA, but I am not able to come up with a counter example. Any help is highly appreciated.

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lkj18

7 лет назад, # ^ |

It doesn't work for some cases where R and B have a large difference. Depending on the order of iteration over the pairs with A+B=2, the case 39 3 might fail for example, choosing

But you could optimally choose (1,1), (2,1), (8,0) instead of (0,2), (11,0).

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7 лет назад, # ^ |

I see. Thank you so much! :)
Just curious, did it come intuitively to you, or you generated some test cases and tested against an AC code?

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lkj18

7 лет назад, # ^ |

← Rev. 2 →

During the contest I at first thought greedy might be correct, but after some time had the intuition that there would probably be a counterexample with larger numbers and larger difference. For actually finding a counterexample now I had to generate some test cases.

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udalov

7 лет назад, # ^ |

11 11 -> 9:

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