Submission #1078299

#	Time	Username	Problem	Language	Result	Execution time	Memory
1078299		c2zi6	Beech Tree (IOI23_beechtree)	C++17	0 / 100	2058 ms	348 KiB

beechtree

#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;
    }
};
namespace TEST3 {
    bool check(VI par) {
        rep(i, par.size()) if (i) {
            if (par[i] != 0 && par[par[i]] != 0) return false;
        }
        return true;
    }
    int n;
    VVI gp;
    VI par;
    VI color;
    set<int> scol(int u) {
        set<int> ret;
        for (int u : gp[u]) ret.insert(color[u]);
        return ret;
    }
    VI solve(int N_ARG, VVI GP_ARG, VI PAR_ARG, VI COLOR_ARG) {
        n = N_ARG;
        gp = GP_ARG;
        par = PAR_ARG;
        color = COLOR_ARG;
        VI answer(n, 1);
        answer[0] = false;
        replr(u, 1, n-1) {
            if (scol(u).size() != gp[u].size()) answer[u] = false;
        }
        set<int> allowed = scol(0);
        if (allowed.size() != gp[0].size()) {
            return answer;
        }
        replr(u, 1, n-1) if (allowed.count(color[u]) == 0) {
            return answer;
        }
        map<int, VI> cnt;
        for (int u : gp[0]) {
            for (int v : gp[u]) {
                cnt[color[v]].pb(u);
            }
        }
        VVI a;
        for (auto& p : cnt) a.pb(p.ss);
        sort(all(a), [&](const VI& a, const VI& b){return a.size() < b.size();});
        VI construct;
        VI ka(n);
        for (VI& vec : a) {
            for (int x : vec) {
                if (ka[x]) continue;
                construct.pb(x);
                ka[x] = true;
            }
            if (construct.size() > vec.size()) {
                return answer;
            }
        }
        answer[0] = true;
        return answer;
    }
};
namespace TEST4 {
    int n;
    VVI gp;
    VI par;
    VI color;
    set<int> scol(int u) {
        set<int> ret;
        for (int u : gp[u]) ret.insert(color[u]);
        return ret;
    }
    bool getans(int u) {
        int mec = -1;
        set<int> my = scol(u);
        if (my.size() != gp[u].size()) return false;
        for (int v : gp[u]) {
            if (gp[v].size()) {
                if (mec == -1) mec = v;
                else return false;
            }
        }
        if (mec == -1) return true;
        // mec-i guyner@ pti exnin u-i guyneri entabazmutyun
        for (int v : gp[mec]) {
            if (my.count(color[v]) == 0) return false;
        }
        return true;
    }
    VI solve(int N_ARG, VVI GP_ARG, VI PAR_ARG, VI COLOR_ARG) {
        n = N_ARG;
        gp = GP_ARG;
        par = PAR_ARG;
        color = COLOR_ARG;
        VI ans(n);
        rep(u, n) ans[u] = getans(u);
        return ans;
    }
};

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);

    /*if (TEST2::check(P)) return TEST2::solve(C);*/
    /*if (n > 8) return TEST3::solve(n, gp, par, color);*/

    answer = VI(n);
    dfs();

    VI ans1 = TEST4::solve(n, gp, par, color);

    for (int x : ans1) cout << x << " "; cout << endl;
    return answer;
}

Compilation message (stderr)

beechtree.cpp: In function 'VI beechtree(int, int, VI, VI)':
beechtree.cpp:194:5: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
  194 |     for (int x : ans1) cout << x << " "; cout << endl;
      |     ^~~
beechtree.cpp:194:42: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
  194 |     for (int x : ans1) cout << x << " "; cout << endl;
      |                                          ^~~~

• Subtask 0: 0 / 0
• Subtask 1: 0 / 9
• Subtask 2: 0 / 5
• Subtask 3: 0 / 9
• Subtask 4: 0 / 8
• Subtask 5: 0 / 14
• Subtask 6: 0 / 14
• Subtask 7: 0 / 12
• Subtask 8: 0 / 17
• Subtask 9: 0 / 12