Submission #1077986

#TimeUsernameProblemLanguageResultExecution timeMemory
1077986c2zi6Beech Tree (IOI23_beechtree)C++17
9 / 100
2086 ms44372 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"

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) {
    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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