Submission #1078005

#TimeUsernameProblemLanguageResultExecution timeMemory
1078005c2zi6Beech Tree (IOI23_beechtree)C++17
14 / 100
2084 ms7760 KiB
#define _USE_MATH_DEFINES #include <bits/stdc++.h> #define ff first #define ss second #define pb push_back #define all(a) (a).begin(), (a).end() #define replr(i, a, b) for (int i = int(a); i <= int(b); ++i) #define reprl(i, a, b) for (int i = int(a); i >= int(b); --i) #define rep(i, n) for (int i = 0; i < int(n); ++i) #define mkp(a, b) make_pair(a, b) using namespace std; typedef long long ll; typedef long double ld; typedef pair<int, int> PII; typedef vector<int> VI; typedef vector<PII> VPI; typedef vector<VI> VVI; typedef vector<VVI> VVVI; typedef vector<VPI> VVPI; typedef pair<ll, ll> PLL; typedef vector<ll> VL; typedef vector<PLL> VPL; typedef vector<VL> VVL; typedef vector<VVL> VVVL; typedef vector<VPL> VVPL; template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;} template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;} #include <ext/pb_ds/assoc_container.hpp> using namespace __gnu_pbds; template<class T> using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; #include "beechtree.h" namespace TEST2 { bool check(VI par) { rep(i, par.size()) if (par[i] != i-1) return false; return true; } VI solve(VI color) { int n = color.size(); VI answer(n); bool ans = true; reprl(i, n-1, 0) { answer[i] = ans; if (color[i] != color[n-1]) ans = false; } return answer; } }; int n; VVI gp; VI par; VI color; VI answer; VI dfs(int u = 0) { VI ret; for (int v : gp[u]) { VI his = dfs(v); for (int x : his) ret.pb(x); } bool ardyoq = false; ret.pb(u); sort(all(ret)); do { if (ret[0] != u) continue; VI prev; rep(i, ret.size()) if (i) { int u = ret[i]; int cnt = 0; for (int v : prev) if (color[v] == color[u]) cnt++; if (ret[cnt] != par[u]) goto vat; prev.pb(u); } ardyoq = true; break; vat:; } while (next_permutation(all(ret))); answer[u] = ardyoq; return ret; } VI beechtree(int N, int M, VI P, VI C) { if (TEST2::check(P)) return TEST2::solve(C); n = N; par = P; color = C; gp = VVI(n); rep(u, n) if (par[u] != -1) gp[par[u]].pb(u); answer = VI(n); dfs(); return answer; }
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