### mango_lassi's blog

By mango_lassi, history, 19 months ago,

The 2019 NWERC (Northwestern European regional contest) is tomorrow. You should see the scoreboard here once the contest starts.

There will be an online mirror here.

Let's discuss the problems after the contest!

• +38

 » 19 months ago, # |   +13 The contest starts in 20 minutes, the online mirror starts in 50.There will also be a live stream on ICPC Live
 » 19 months ago, # |   +16 How to solve B, D, H, J and K?
•  » » 19 months ago, # ^ |   +11 J: Make a segment tree that tells us for every interval how many removals we need to make to make that interval of posts into a alternating sequence that starts with negative / positive and ends with negative/positive. Combining this information is trivial.Initially each post must either be deleted or have the sign it initially has. Loop the amount of new accounts we create, when the number of accounts goes over $abs(votes_i)$, that position in the segment tree can now have any sign, so we update it. Every position gets updated at most once, and the segment tree operations work in $O(\log n)$, so that's our time complexity.
•  » » 19 months ago, # ^ |   +21 D: For each edge count, find the shortest path with exactly this many edges when ignoring the constant x that is added to each edge. The actual length of such a path is then a line equation in terms of x. You then need to find all the lines which are minimal for at least one value (kind of like in the convex hull trick). Finally you need to find all the vertices which are on at least one of these paths.
•  » » 19 months ago, # ^ |   +21 You can also see the solution presentation if you skip back in the live broadcast.
 » 19 months ago, # |   +8 Did anybody manage to solve B?
•  » » 19 months ago, # ^ | ← Rev. 2 →   +8 Idk if my solution is correct, but I did manage to squeeze it in 4s solution/*input 8 5 3 1 -1 5 7 5 3 7 */ #pragma GCC optimize ("O3") #include using namespace std; typedef long long ll; typedef long double ld; const int mxh = 27; vector> dp[505050][mxh]; vectorch[505050]; int pa[505050]; int h[505050]; int sz[505050]; vector>suma( vector> & a, vector> & b) { if (a.empty() || b.empty()) return {}; vector>c; if (a.size() + b.size() >= 3) { return {{a[0].first + b[0].first, a.back().second + b.back().second}}; } int x = 1e8; int y = -1; for (pairi : a) { for (pairj : b) { x = min(x, i.first + j.first); y = max(y, i.second + j.second); } } return {{x, y}}; //sort(c.begin(), c.end()); //return fix(c); } vector>pluss( vector> &a) { vector>r; for (pairi : a) { r.push_back({i.first + 1, i.second + 1}); } return r; } void merge( vector> & a, vector> & b, vector> & c, vector> & r) { if (a.size() + b.size() + c.size() >= 4) { int x = 1e8; int y = -1; if (!a.empty()) { x = min(x, a[0].first); y = max(y, a.back().second); } if (!b.empty()) { x = min(x, b[0].first); y = max(y, b.back().second); } if (!c.empty()) { x = min(x, c[0].first); y = max(y, c.back().second); } r = {{x, y}}; return; } r.clear(); int i = 0; int j = 0; int k = 0; int d = -100; auto add = [&](pairi) { if (i.first <= d) { r.back().second = max(r.back().second, i.second + 1); } else { r.push_back({i.first + 1, i.second + 1}); } d = r.back().second; }; while (i < a.size() && j < b.size() && k < c.size()) { if (a[i] <= b[j]) { if (a[i] <= c[k]) { add(a[i++]); } else { add(c[k++]); } } else { if (b[j] <= c[k]) { add(b[j++]); } else { add(c[k++]); } } } while (j < b.size() && k < c.size()) { if (b[j] < c[k]) { add(b[j++]); } else { add(c[k++]); } } while (i < a.size() && k < c.size()) { if (a[i] < c[k]) { add(a[i++]); } else { add(c[k++]); } } while (i < a.size() && j < b.size()) { if (a[i] < b[j]) { add(a[i++]); } else { add(b[j++]); } } while (i < a.size()) add(a[i++]); while (j < b.size()) add(b[j++]); while (k < c.size()) add(c[k++]); } void prec(int v) { sz[v] = 1; h[v] = 1; for (int w : ch[v]) { prec(w); sz[v] += sz[w]; h[v] = max(h[v], h[w] + 1); } dp[v][0] = {{0, 0}}; dp[v][1] = {{1, 1}}; if (ch[v].size() == 1) { dp[v][2] = {{2, 2}}; } if (ch[v].size() == 2) { int &l = ch[v][0]; int &r = ch[v][1]; if (l > r) swap(l, r); for (int i = 2; i <= h[v]; i++) { auto v1 = suma(dp[l][i - 1], dp[r][i - 1]); auto v2 = suma(dp[l][i - 1], dp[r][i - 2]); auto v3 = suma(dp[l][i - 2], dp[r][i - 1]); merge( v1, v2, v3, dp[v][i] ); } } } int K[505050]; int NE[505050]; int n; int k; int r = 1; void blogas(int i) { if (NE[i]) return; NE[i] = 1; for (int j : ch[i]) { blogas(j); } } void check(int i) { if (K[i] == 1 || NE[i] == 1) return; vectornauji; vectorcheck; int j = i; while (j != -1) { if (K[j] == 0) nauji.push_back(j); check.push_back(j); j = pa[j]; } for (int x : nauji) { K[x] = 1; } for (int v : check) { for (int i = 0; i < mxh; i++) { dp[v][i] = {}; } if (ch[v].size() == 0) { dp[v][1] = {{1, 1}}; } if (ch[v].size() == 1) { int w = ch[v][0]; dp[v][1] = pluss(dp[w][0]); dp[v][2] = pluss(dp[w][1]); } if (ch[v].size() == 2) { int l = ch[v][0]; int r = ch[v][1]; auto v4 = suma(dp[l][0], dp[r][0]); dp[v][1] = pluss(v4); for (int i = 2; i <= h[v]; i++) { auto v1 = suma(dp[l][i - 1], dp[r][i - 1]); auto v2 = suma(dp[l][i - 1], dp[r][i - 2]); auto v3 = suma(dp[l][i - 2], dp[r][i - 1]); merge( v1, v2, v3, dp[v][i] ); } } } bool ok = false; for (int h = 0; h < mxh; h++) for (paira : dp[r][h]) if (a.first <= k && k <= a.second) ok = true; if (ok) return; for (int x : nauji) { K[x] = 0; } for (int v : check) { for (int i = 0; i < mxh; i++) { dp[v][i] = {}; } if (K[v] == 0) { dp[v][0] = {{0, 0}}; } else { if (ch[v].size() == 0) { dp[v][1] = {{1, 1}}; } if (ch[v].size() == 1) { int w = ch[v][0]; dp[v][1] = pluss(dp[w][0]); } if (ch[v].size() == 2) { int l = ch[v][0]; int r = ch[v][1]; auto v4 = suma(dp[l][0], dp[r][0]); dp[v][1] = pluss(v4); for (int i = 2; i <= h[v]; i++) { auto v1 = suma(dp[l][i - 1], dp[r][i - 1]); auto v2 = suma(dp[l][i - 1], dp[r][i - 2]); auto v3 = suma(dp[l][i - 2], dp[r][i - 1]); merge( v1, v2, v3, dp[v][i] ); } } } } blogas(i); } int main() { ios_base::sync_with_stdio(false); cin >> n >> k; for (int i = 1; i <= n; i++) { cin >> pa[i]; if (pa[i] == -1) r = i; else ch[pa[i]].push_back(i); } prec(r); for (int i = 1; i <= n; i++) { check(i); } for (int i = 1; i <= n; i++) { cout << "01"[K[i]]; } } I maintained all possible subtree sizes with a specific height by a vector of intervals. To get lexicographically smallest subtree I just checked if a specific prefix is possible. This needs some nasty optimizations because in almost all cases dp value is just one interval, but not always. Can anyone find a counter test for my solution?
•  » » 19 months ago, # ^ | ← Rev. 2 →   +18 I tried to write $O(n\log n)$ and got WA, but $O(n\log^2n)$ passes easily anyways. Designate a set of nodes such that none of them may be erased, and try adding nodes to this set in DFS order. After adding a node to the set, test whether the minimum possible size of a BBST containing all nodes in the set has size at most $k;$ if not, remove the node from the set.CodeUPDATE: Here's $O(n\log n)$ (apparently it's slower tho lol)Code
•  » » » 19 months ago, # ^ |   +8 It's also possible in linear time.
 » 19 months ago, # |   +26 What did happen with Oxford team ? They had 5 harder before frozen, and currently they are not on the leaderboard.
•  » » 19 months ago, # ^ |   0 I'm not exactly sure, but they were not finals-eligible. Maybe they got disqualified
•  » » » 19 months ago, # ^ | ← Rev. 2 →   +18 .
 » 19 months ago, # |   +24 Will this contest be put into gym/opentrains for virtual participation?
•  » » 19 months ago, # ^ |   +10 In theory anyone can do it. The problem packages are available on the site: http://www.nwerc.eu/ -- most of the jury are travelling back from NWERC to our normal lives today. If nobody else has added the packages to Codeforces by the weekend, I can do it.You can also find the questions on Kattis already because like usual they hosted the semilive contest for us: Problems from Northwestern Europe Regional Contest