Submission #915898

#	Time	Username	Problem	Language	Result	Execution time	Memory
915898		EJIC_B_KEDAX	Group Photo (JOI21_ho_t3)	C++17	100 / 100	333 ms	196688 KiB

Main

#ifdef LOCAL
    # define _GLIBCXX_DEBUG
#endif
#include <bits/stdc++.h>
#include <immintrin.h>

#ifndef LOCAL
    // #pragma GCC optimize("O3")
    // #pragma GCC optimize("Ofast")
    // #pragma GCC optimize("unroll-loops")
    // #pragma GCC target("avx,avx2,bmi,bmi2,popcnt,lzcnt,sse,sse2,sse3,ssse3,mmx")
#endif
using namespace std;
using ll = long long;
using ld = long double;
# define x first
# define y second
# define all(x) x.begin(), x.end()
# define rall(x) x.rbegin(), x.rend()

mt19937_64 mt(time(0));

void solve();
void init();

int32_t main() {
#ifndef LOCAL
    cin.tie(nullptr)->sync_with_stdio(false);
#endif
    cout << fixed << setprecision(30);
    init();
    int t = 1;
    // cin >> t;
    while (t--) {
        solve();
    }
}

void init() {}

void solve() {
    int n;
    cin >> n;
    vector<pair<int, int>> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i].x;
        a[i].y = i;
    }
    sort(all(a), greater<>());
    vector<vector<int>> dp1(n, vector<int>(n)), dp2(n, vector<int>(n));
    for (int i = 0; i < n; i++) {
        dp2[i][i] = 0;
        for (int j = i - 1; j >= 0; j--) {
            if (a[j].y > a[i].y) {
                dp2[i][j] = dp2[i][j + 1] + 1;
            } else {
                dp2[i][j] = dp2[i][j + 1];
            }
        }
    }
    for (int i = 0; i < n; i++) {
        dp1[i][i] = 0;
        for (int j = i + 1; j < n; j++) {
            dp1[i][j] = dp1[i][j - 1] + dp2[j][i];
        }
    }
    int off = n * (n - 1) / 2 - dp1[0][n - 1];
    vector<int> dp(n + 1, 0);
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            dp1[i][j] = 2 * dp1[i][j] - (j - i) * (j - i + 1) / 2;
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = i - 1; j >= 0; j--) {
            dp[i] = min(dp[i], dp[j] + dp1[j][i - 1]);
        }
    }
    cout << off + dp[n] << '\n';
}

• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 32 / 32
• Subtask 4: 20 / 20
• Subtask 5: 36 / 36