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Discussion of Problem 1022. Genealogical Tree

accepted
Posted by Mikhail 16 May 2018 12:07
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("avx")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#define re return
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define sqrt(x) sqrt(abs(x))
#define mp make_pair
#define pi (3.14159265358979323846264338327950288419716939937510)
#define fo(i, n) for(int i = 0; i < n; ++i)
#define ro(i, n) for(int i = n - 1; i >= 0; --i)
#define unique(v) v.resize(unique(all(v)) - v.begin())

template <class T> T abs (T x) { re x > 0 ? x : -x; }
template <class T> T sqr (T x) { re x * x; }
template <class T> T gcd (T a, T b) { re a ? gcd (b % a, a) : b; }
template <class T> int sgn (T x) { re x > 0 ? 1 : (x < 0 ? -1 : 0); }

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef double D;
typedef long double ld;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef unsigned long long ull;
typedef tree <pair<int, char>, null_type, less<pair<int, char>>, rb_tree_tag, tree_order_statistics_node_update> _tree;

const int maxn = 100;
vi g[maxn], rg[maxn];
int ans[maxn];
bool was[maxn];

int dfs (int v, int time) {
    //cout << v + 1 << ' ' << time << endl;
    int start_time = time;
    was[v] = true;
    for (int j : rg[v]) {
        if (!was[j]) time += dfs(j, time++);
    }
    ans[v] = time;
    for (int j : g[v]) {
        if (!was[j]) time += dfs(j, ++time);
    }
    re time - start_time;
}

int main() {
    int n;
    cin >> n;
    int x;
    fo(i, n) {
        cin >> x;
        while (x) {
            g[i].pb(x - 1);
            rg[x - 1].pb(i);
            cin >> x;
        }
    }
    int time = 0;
    fo(i, n) {
        if (!was[i]) time += dfs(i, time);
    }
    fo(i, n) {
        fo(j, n) if (ans[j] == i) cout << j + 1 << ' ';
    }
    cout << endl;
    //fo(i, n) cout << ans[i] << ' ';
    re 0;
}