Submission #685557

#	Time	Username	Problem	Language	Result	Execution time	Memory
685557		stanislavpolyn	Village (BOI20_village)	C++17	0 / 100	1084 ms	5272 KiB

Village

#include <bits/stdc++.h>

#define fr(i, a, b) for (int i = (a); i <= (b); ++i)
#define rf(i, a, b) for (int i = (a); i >= (b); --i)
#define fe(x, y) for (auto& x : y)

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define all(x) (x).begin(), (x).end()
#define sz(x) (int)(x).size()
#define pw(x) (1LL << (x))

using namespace std;

mt19937_64 rng(228);

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <typename T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define fbo find_by_order
#define ook order_of_key

template <typename T>
bool umn(T& a, T b) {
    return a > b ? a = b, 1 : 0;
}
template <typename T>
bool umx(T& a, T b) {
    return a < b ? a = b, 1 : 0;
}

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
using ve = vector<T>;

const int N = 1e5 + 5;

int n;
ve<int> g[N];

int dist[1001][1001];
int start;

void dfs(int v, int p, int dep) {
    dist[start][v] = dep;
    fe (to, g[v]) {
        if (to == p) continue;
        dfs(to, v, dep + 1);
    }
}

int p[N];
ve<int> order;

void dfs1(int v = 1, int p = 0) {
    ::p[v] = p;
    fe (to, g[v]) {
        if (to == p) continue;
        dfs1(to, v);
    }
    order.pb(v);
}

void assertTL(bool b) {
    if (!b) {
        while (1);
    }
}

int main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#else
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif
    cin >> n;

    fr (i, 1, n - 1) {
        int a, b;
        cin >> a >> b;

        g[a].pb(b);
        g[b].pb(a);
    }

    if (n <= 1000) {
        for (start = 1; start <= n; start++) {
            dfs(start, 0, 0);
        }
    }

    dfs1();

    ll mySum = -1;
    {
        ve<bool> used(n + 1);
        ve<int> to(n + 1);

        fe (v, order) {
            ve<int> sons;
            fe (to, g[v]) {
                if (to == p[v]) continue;

                if (!used[to]) {
                    sons.pb(to);
                }
            }

            while (sz(sons) > 1) {
                used[sons.back()] = 1;
                used[sons[sz(sons) - 2]] = 1;
                to[sons.back()] = sons[sz(sons) - 2];
                to[sons[sz(sons) - 2]] = sons.back();
                sons.pop_back();
                sons.pop_back();
            }

            if (sz(sons)) {
                used[sons.back()] = 1;
                used[v] = 1;
                to[sons.back()] = v;
                to[v] = sons.back();
                sons.pop_back();
            }
        }
        if (n % 2) {
            assert(!to[1]);
            bool found = 0;

            fe (u, g[1]) {
                if (p[to[u]] == 1) {
                    int v = to[u];
                    to[v] = 1;
                    to[1] = u;
                    to[u] = v;

                    found = 1;
                    break;
                }
            }

            if (!found) {
                fe (u, g[1]) {
                    int v = to[u];
                    to[1] = u;
                    to[u] = v;
                    to[v] = 1;
                    break;
                }
            }
        }

        ll sum = 0;
        fr (i, 1, n) {
            sum += dist[i][to[i]];
        }


        set<int> s;
        fr (i, 1, n) {
            s.insert(to[i]);
            assertTL(to[i] != i);
        }
        assertTL(sz(s) == n);

        mySum = sum;
//        cout << sum << "\n";
//        fr (i, 1, n) {
//            cout << to[i] << " ";
//        }
//        cout << "\n";
    }

    {
        ll mx = -1e18;
        ve<int> mxP;
        ll mn = 1e18;
        ve<int> mnP;

        ve<int> p;
        fr (i, 1, n) p.pb(i);
        do {
            bool bad = 0;
            fr (i, 0, sz(p) - 1) {
                bad |= p[i] == i + 1;
            }
            if (!bad) {
                ll sum = 0;
                fr (i, 0, sz(p) - 1) {
                    sum += dist[i + 1][p[i]];
                }
                if (umx(mx, sum)) {
                    mxP = p;
                }

                bad = 0;

                if (!bad) {
                    if (umn(mn, sum)) {
                        mnP = p;
                    }
                }
            }
        } while (next_permutation(all(p)));

        assert(mySum == mn);

        cout << mn << " " << mx << "\n";
        fe (x, mnP) cout << x << " ";
        cout << "\n";

        fr (i, 0, sz(mnP) - 1) {
            assert(dist[i + 1][mnP[i]] <= 2);
        }
        fe (x, mxP) cout << x << " ";
        cout << "\n";
        cout << "\n";
    }

    return 0;
}

• Subtask 1: 0 / 12
• Subtask 2: 0 / 38
• Subtask 3: 0 / 50