제출 #1244616

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1244616		M_SH_O	Group Photo (JOI21_ho_t3)	C++20	100 / 100	930 ms	393396 KiB

Main

/*#pragma GCC optimize("O3")
#pragma GCC optimization("Ofast,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")*/
#include <bits/stdc++.h>
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>*/

#define ll long long
#define ll1 long long
#define ull unsigned long long
#define dou long double
#define str string
#define vll vector<ll>
#define vi vector<int>
#define pll pair<ll, ll>
#define vpll vector<pll>
#define vbool vector<bool>
#define vstr vector<str>
#define vvll vector<vll>
#define pb push_back
#define pf push_front
#define endl "\n"
#define fr first
#define se second
// #define sortcmp(a) sort(a.begin(), a.end(), cmp)
#define sort(a) sort(a.begin(), a.end())
#define reverse(a) reverse(a.begin(), a.end())
#define speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define INF 1000000000000000007
#define ordered_set tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update>

using namespace std;
//using namespace __gnu_pbds;

mt19937 rng(time(0));
ll randll(ll l, ll r) {
    return uniform_int_distribution<ll>(l, r)(rng);
}

int main(){
    speed;
    srand(time(0));

    ll n;
    cin >> n;
    vll a(n), b(n+7);
    for (int i =0 ; i < n; i ++) {
        cin >> a[i];
        b[a[i]] = i;
    }

    vll dp(n+7, INF);
    dp[0] = 0;

    vvll p(n+7, vll(n+7, 0)), pr(n+7, vll(n+7, 0));
    vbool us(n+7, 0);
    for (int i = 0; i < n; i ++) {
        us[a[i]] = 1;
        for (int j = 1; j <= n; j ++) {
            p[i][j] = p[i][j-1]+us[j];
        }
    }

    us.clear();
    us.resize(n+7, 0);
    for (int i = n-1; i >= 0; i --) {
        us[a[i]] = 1;
        for (int j = 1; j <= n; j ++) {
            pr[i][j] = pr[i][j-1]+us[j];
        }
    }

    for (int i = 1; i <= n; i ++) {
        ll res = 0;
        for (int j = i; j >= 1; j --) {

            ll v = b[j];
            res += p[v][n]-p[v][i];
            res += pr[v][i]-pr[v][j];

            dp[i] = min(dp[i], dp[j-1]+res);
        }
    }

    cout << dp[n] << endl;
}

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