제출 #593766

#제출 시각아이디문제언어결과실행 시간메모리
593766bogdanvladmihaiBall Machine (BOI13_ballmachine)C++14
16.11 / 100
148 ms33292 KiB
#include <bits/stdc++.h> using namespace std; /*** Given a rooted tree, you have to support 2 types of operations: * paint the root of a full black subtree in black * check where the path from a black node to the root becomes white I can do the erases using binary lifting: * if the 2^k ancestor of node u is black, than all the nodes between them are black * if not, you can go down the tree Add operation: * I will always want to add in the same exact order * So I will have a vector of 1 and 0, and I want to find the LAST 0 ***/ const int LG = 17; const int MAXN = 1000 * 100; vector<int> g[MAXN + 1], order[MAXN + 1]; int dad[MAXN + 1], minValue[MAXN + 1], who[MAXN + 1], tin[MAXN + 1]; int ancestor[LG + 1][MAXN + 1]; bool black[MAXN + 1]; priority_queue<int> fr; void dfs(int u) { vector<pair<int, int>> children; for (const int &v : g[u]) { dfs(v); minValue[u] = min(minValue[u], minValue[v]); children.emplace_back(minValue[v], v); } sort(children.begin(), children.end()); for (const auto &p : children) { order[u].push_back(p.second); } } void dfsOrder(int u) { static int t = 0; tin[u] = t++; who[t - 1] = u; for (const int &v : order[u]) { dfsOrder(v); } } int add() { int u = who[fr.top()]; fr.pop(); black[u] = true; return u; } int erase(int u) { int ans = 0; for (int l = LG; l >= 0; l--) { if (black[ancestor[l][u]]) { u = ancestor[l][u]; ans += (1 << l); } } fr.push(tin[u]); black[u] = false; return ans; } int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); int n, q; cin >> n >> q; int root = -1; for (int i = 1; i <= n; i++) { cin >> dad[i]; if (dad[i] != 0) { g[dad[i]].push_back(i); } else { root = i; } } for (int i = 1; i <= n; i++) { minValue[i] = i; fr.push(i - 1); } dfs(root); dfsOrder(root); for (int u = 1; u <= n; u++) { ancestor[0][u] = dad[u]; } for (int k = 1; k <= LG; k++) { for (int u = 1; u <= n; u++) { ancestor[k][u] = ancestor[k - 1][ancestor[k - 1][u]]; } } for (int i = 0; i < q; i++) { int op, k; cin >> op >> k; if (op == 1) { int u = -1; for (int j = 0; j < k; j++) { u = add(); } cout << u << "\n"; } else { cout << erase(k) << "\n"; } } return 0; }
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