lukasP's blog

NWERC 2014
By lukasP, 9 years ago, In English

Are you coming to NWERC? Please comment in the TopCoder Forums.

There will also be an online contest. The contest starts on Sunday 2014-11-30 at 11:00 AM CET and lasts for 5 hours.

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9 years ago, # |
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There will be an online contest. I have updated my post above with the correct URL and start time.

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    9 years ago, # ^ |
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    Reminder: the online contest starts tomorrow. You can also watch the onsite contest live as well as watch the live feed.

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      9 years ago, # ^ |
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      Is it correct that the online contest starts one hour later than the actual contest?

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        9 years ago, # ^ |
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        Yes.

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          9 years ago, # ^ |
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          Hm, so online participants will have an advantage, because they will see what is done in "one hour in future". But on the other side since online contest is (mainly) individual then it shouldn't be that much easier :D.

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            9 years ago, # ^ |
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            Yes, but the online contest is a lower priority for us and for technical reasons starting both at the same time is impossible.

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9 years ago, # |
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a very good contest,i will take part in

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9 years ago, # |
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Is it possible to create a team for online contest? If yes, please say how.

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    9 years ago, # ^ |
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    Yes, but it's a manual process. Please send me your team name and logins of the team members.

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9 years ago, # |
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Will ranking in online contest include teams from onsite contest? Or will we have access to ranking of onsite teams at least?

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    9 years ago, # ^ |
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    The rankings will be separate, but you can watch the onsite contest here.

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9 years ago, # |
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The online contest starts in about two hours!

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9 years ago, # |
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How to solve problem F, I , K ?

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    9 years ago, # ^ |
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    I: meet in the middle on half-cycles, watch out for constants

    I suppose F is divide and conquer while bounding the number of interesting lines in a subset of points. I'm not sure what kind of bounding works, though. (UPD: Nah, I'm probably just getting the due to counting the same line many times.)

    K seemed like just efficient simulation when starting at any knapsack + computing sums for all positions from that (the time should increase linearly when moving away from a knapsack till encountering another one)... not sure, I didn't try to code it.

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      9 years ago, # ^ |
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      I thought about the Meet in the middle technique but could not figure it out for problem I. Can you explain your idea ?

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        9 years ago, # ^ |
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        Edit: I explained F instead :D

        Try all choices for the first N / 2 vertices of any path and remember (in a set) just the starting, ending vertex, set of vertices in the path and length of the path. The number of choices is . Backtrack, bitmasking, anything works (but the time limit is a bit tight).

        Suppose that you have two halves of a path: one between u1 and v1 and containing vertices from a set S and with length l1, the other between u2 and v2 containing all vertices not in S. Try all choices of u1, v1, u2, v2, S, l1. Try to finalize the cycle by connecting u1, v1 and u2, v2; now, the cost of the second half-path is uniquely given, so you can just check if it exists in the corresponding set.

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          9 years ago, # ^ |
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          He asked about problem I, not F ; p.

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            9 years ago, # ^ |
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            Whoops. The principles are so similar that I didn't even notice.

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      9 years ago, # ^ |
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      In F p is always > 20. If line that contais > p% of points exists, then pair of randomly choosen points lies on it with probality > 0.04. So we can pick random pair of points several hundred times and either find answer or find out that it doesn't exists.

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        9 years ago, # ^ |
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        Oh, of course!

        I'm not really used to trying to find something like a randomized algorithm. I think the divide and conquer should work, but my code is getting WA somewhere. I wonder when the tests will be public...

        UPD: solved it using divide and conquer, I just missed the case when N = 1 *facepalm*

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        9 years ago, # ^ |
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        Why a random pair of points? Just pick a single point, which has a 20% chance of being in the line. Count up the number of other points in each direction to check it.

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9 years ago, # |
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How to solve A?

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9 years ago, # |
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Though easy problems were pretty ok, I didn't enjoy problemset in general, because 3 hardest problems were geos, no problem demanded any deeper insight (I think I figured out tricky cases in A and B, though in geo you never know) and problem "I" was really bad — I came up with good solution really quickly, but it was hard for me to squeeze it in TL, I needed 11 attempts and they differed by small optimizations.

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    9 years ago, # ^ |
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    On the other hand, I think that NCPC problems were really nice :).

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    9 years ago, # ^ |
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    I also made 6 attempts (though 2 of them were WA 1 -_-). In my last optimization, I realized that since it is a cycle, we can fix 1 point at 0, thus gain 7x speed up. My last submission runs in 2.25s while TL is 10s

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    9 years ago, # ^ |
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    For I there are plenty of solutions that passed well below the time limit. Your approach was probably not the fastest one.

    In problem A one can come up with many "deeper inisghts" that simplify the coding. And I would also disagree on whether G was a geometry or not. It's Manhattan metric after all.

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      9 years ago, # ^ |
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      I had complexity in problem I. Is there something significantly faster in terms of complexity?

      And talking about A I agree that one probably should think much before starting coding it. I started, but I quickly realized that my approach is completely wrong, because I wasn't taking care of doing a lap, which was in fact main point here. I think that we should draw two half lines with equation x = x1, y ≥ y1 and x = x1, y ≤ y1 and consider crossing them in an appropriate order (of course creating a graph of admissible connections between them and doing some dp/Dijkstra). Is that right? Even if it is then I should think about many coding issues :P. And I have to admit that this problem is really taken from "real life", which is always a plus :).

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        9 years ago, # ^ |
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        I did a Dijkstra but I was calculating winding number instead. Try solving this problem first and then modify that approach for A. Per Austrin thinks his convex hull solution is nice too.

        The slides are up now, you can check your I solution.

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          9 years ago, # ^ |
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          Hah, ternary search was essential/useful in 3 tasks (B, E, G) :D. And in B it was nested ternary search :P (after seeing the editorial I have to admit that I definitely didn't consider all cases and haven't thought about ternary :P). It reminded me of a problem when I ternary searched in ternary search in order to find place where is the best result and when fixing parameters I computed result in that place using binary search :P (problem B here: http://codeforces.com/gym/100491).

          And model solution to I was essentialy the same as mine and resulted in exactly the same complexity. I would even say that my solution is more elegant, because it doesn't middle vertex and do that whole partitioning: http://ideone.com/cuWbcC It is basically Dfs considering all possible paths of length <=n/2.

          Btw sorry that in previous post I asked about is there something faster than complexity which was an expression which couldn't be parsed xD. Codeforces got down for a while, so I didn't do a preview, now it's ok :pp.

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            9 years ago, # ^ |
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            E? Pfft, binsearch on the derivative.

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              9 years ago, # ^ |
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              Why? This is not a big difference between ternary and binary search and you do not have to do some not that nice computations about derivatives/any further thinking whether derivative is positive/negative.

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                9 years ago, # ^ |
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                Why? Why not? :D

                It's just that I have better skill in calculus than in coding ternary search, and if I express the original function as , with simple formulas for K, E, L, then I see immediately that the derivative is — and thinking about signs is a matter of seconds (large e: positive derivative, small e: negative).

                I just pointed it out as "ha! but I'm different!" :D

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How to solve C and E?

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9 years ago, # |
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How to solve G?

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    9 years ago, # ^ |
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    I didn't code it, but I think this should work:

    It is trivial, that if d = inf, then x *  = med(x1, ..., xn), where med is a median function and analogous property holds for y * .

    Firstly create circles with radius d for all points from input as their centers (in Manhattan metric, so in fact these are squares with sides parallel/perpendicular to y=x) and intersect them all, it will be a rectangle. If (med(X), med(Y)) lies within that field than we're done. If not, we should check all 4 sides separately (or even just 1 or 2, but doesn't matter). Each side can be checked in O(n) if we will divide it into intervals where we are not crossing x or y coordinate of any of the points.

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      9 years ago, # ^ |
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      It's easier to do last step with ternary search by x.

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        9 years ago, # ^ |
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        Good point, it surely simplifies code :) (this is not an irony, to be clear :p).

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    9 years ago, # ^ |
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    You can also find rectangle in rotated plane in which solution must lie and then do 2D ternary search on it. You just have to be careful that not all points in that rectangle map to integer points in normal plane (only those with x%2==y%2 do), easiest thing to do is to do search twice, once for odd points and once for even points.

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Videos of problem analysis.

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How many teams will advance to WF from this region?

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6 years ago, # |
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Is there something wrong with problem H for the system?

http://codeforces.com/gym/101482/standings

The following simple code is not passing (WA on test 1). Am I missing something?

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int (i) = 0; (i) < (n); (i) ++)
#define rep1(i, n) for (int (i) = 1; (i) <= (n); (i) ++)
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); (i)++)
#define db(x) {cout << #x << " = " << (x) << endl;}
#define dba(a, x, y) {cout<<#a<<" :";FOR(i123,x,y)cout<<setw(4)<<(a)[i123];cout<<endl;}
#define clr(x) memset(x,0,sizeof(x));
#define mp make_pair
#define pb push_back
#define sz(x) int((x).size())
typedef long long ll;
typedef long double ld;
const int INF = INT_MAX;
const ll INFL = LLONG_MAX;
const ld pi = acos(-1);
// const int MOD = ;

vector<int> G[100100];
pair<int,int> F[100100];
int T = 0;

void dfs(int u, int p, int a){
    if(p == -1){
        F[u].first = ++T;
        F[u].second = ++T;
        rep(i,sz(G[u])){
            int v = G[u][i];
            if(i == 0) dfs(v,u,F[u].first);
            if(i == 1) dfs(v,u,F[u].second);
            if(i >= 2) dfs(v,u,F[u].first);
        }
    }else{
        F[u].first = a;
        F[u].second = ++T;
        for(int v : G[u]){
            if(v == p) continue;
            dfs(v,u,F[u].second);
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(0); cout.precision(15); cout << fixed; cout.tie(0); cin.tie(0);
    int N;
    cin >> N;
    rep(i,N-1){
        int a,b;
        cin >> a >> b;
        G[a].pb(b);
        G[b].pb(a);
    }
    dfs(1,-1,0);
    rep1(i,N) cout << F[i].first << " " << F[i].second << endl;

}
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