Submission #688516

#	Time	Username	Problem	Language	Result	Execution time	Memory
688516		stanislavpolyn	Olympiads (BOI19_olympiads)	C++17	44 / 100	551 ms	17016 KiB

olympiads

#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 pw(x) (1LL << (x))
#define sz(x) (int)(x).size()

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>;

int n, k, c;
int a[505][7];
ve<int> st;
int mx[7];
ve<int> scores;
int sum[505];
ve<int> order;

void rec(int p = 1) {
    if (sz(st) == k) {
        fr (i, 1, k) mx[i] = -1e9;

        fe (cur, st) {
            fr (i, 1, k) {
                umx(mx[i], a[cur][i]);
            }
        }
        int sum = 0;
        fr (i, 1, k) sum += mx[i];

        scores.pb(sum);

        return;
    }
    if (p > n) return;
    rec(p + 1);
    st.pb(p);
    rec(p + 1);
    st.pop_back();
}

int need[7];
int mask[505];

ll dp[7][64];

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

    ve<array<int, 6>> res;

    fr (a1, 0, 10) {
        fr (a2, 0, 10) {
            fr (a3, 0, 10) {
                fr (a4, 0, 10) {
                    fr (a5, 0, 10) {
                        fr (a6, 0, 10) {
                            if (a1 + a2 + a3 + a4 + a5 + a6 <= 10) {
                                res.pb({a1, a2, a3, a4, a5, a6});
                            }
                        }
                    }
                }
            }
        }
    }
//    cout << cnt << "\n";
//
//    return 0;

    cin >> n >> k >> c;

    fr (i, 1, n) {
        fr (j, 1, k) {
            cin >> a[i][j];
        }
    }

    if (n > 40 && k > 2) {
        fe (cur, res) {
            fr (i, k, 5) {
                cur[i] = 0;
            }
        }

        sort(all(res));
        res.erase(unique(all(res)), res.end());

        sort(all(res), [](array<int, 6> a, array<int, 6> b) {
            int sum1 = 0;
            fr (i, 0, k - 1) sum1 += a[i];
            int sum2 = 0;
            fr (i, 0, k - 1) sum2 += b[i];
            return sum1 > sum2;
        });

        rf (j, sz(res) - 1, 0) {
            fr (i, 1, k) {
                need[i] = res[j][i - 1];
            }

            ve<int> have;
            fr (i, 1, n) {
                bool bad = 0;
                fr (j, 1, k) {
                    if (a[i][j] > need[j]) {
                        bad = 1;
                        break;
                    }
                }

                if (!bad) {
                    int mask = 0;
                    fr (j, 1, k) {
                        if (a[i][j] == need[j]) {
                            mask |= pw(j - 1);
                        }
                    }

                    have.pb(mask);
                }
            }

            fr (mask, 0, 63) {
                fr (take, 0, k) {
                    dp[take][mask] = 0;
                }
            }

            dp[0][0] = 1;
            fe (cur, have) {
                rf (was, k - 1, 0) {
                    fr (mask, 0, pw(k) - 1) {
                        dp[was + 1][mask | cur] += dp[was][mask];
                        umn(dp[was + 1][mask | cur], 1001LL);
                    }
                }
            }

            c -= dp[k][pw(k) - 1];
            if (c <= 0) {
                int sum = 0;
                fr (i, 1, k) sum += need[i];
                cout << sum << "\n";
                return 0;
            }
        }

        assert(0);

        return 0;
    }

    rec();

    sort(all(scores));
    reverse(all(scores));

    cout << scores[c - 1] << "\n";

    return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 31 / 31
• Subtask 3: 0 / 24
• Subtask 4: 0 / 32