Submission #751772

#	Time	Username	Problem	Language	Result	Execution time	Memory
751772		GrindMachine	Group Photo (JOI21_ho_t3)	C++17	100 / 100	597 ms	196684 KiB

Main

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

template<typename T>
struct fenwick {
    int siz;
    vector<T> tree;

    fenwick(int n) {
        siz = n;
        tree = vector<T>(n + 1);
    }

    int lsb(int x) {
        return x & -x;
    }

    void build(vector<T> &a, int n) {
        for (int i = 1; i <= n; ++i) {
            int par = i + lsb(i);
            tree[i] += a[i];

            if (par <= siz) {
                tree[par] += tree[i];
            }
        }
    }

    void pupd(int i, T v) {
        while (i <= siz) {
            tree[i] += v;
            i += lsb(i);
        }
    }

    T sum(int i) {
        T res = 0;

        while (i) {
            res += tree[i];
            i -= lsb(i);
        }

        return res;
    }

    T query(int l, int r) {
        if (l > r) return 0;
        T res = sum(r) - sum(l - 1);
        return res;
    }
};

void solve(int test_case)
{
    ll n; cin >> n;
    vector<ll> a(n+5);
    rep1(i,n) cin >> a[i];

    vector<ll> pos(n+5);
    rep1(i,n) pos[a[i]] = i;

    ll cost[n+5][n+5];
    memset(cost,0,sizeof cost);

    rep1(l,n){
        // cost(l,r):
        // consider the array after removing guys > r
        // now we want to take all guys [l,r] to end and arrange them in order r,r-1,r-2,...,l
        // cost(l,r) = cost to take to end + cost to rearrange within in end
        // cost to take to end = sum(#of guys smaller than l that appear after pos[i]), for i = [l..r]
        // cost to rearrange = #of inversions if we consider guys [l..r]
        // our final order must be r,r-1,...,l (decreasing order)
        // so definition of inversions here = #of (i,j) s.t i < j and a[i] < a[j]
    
        vector<ll> suff(n+5);
        rep1(i,n){
            if(a[i] < l){
                suff[i] = 1;
            }
        }

        rev(i,n,1){
            suff[i] += suff[i+1];
        }

        fenwick<ll> fenw(n+5);
        ll take_to_end = 0;
        ll inv = 0;

        for(int r = l; r <= n; ++r){
            take_to_end += suff[pos[r]+1];
            inv += fenw.query(1,pos[r]-1);
            fenw.pupd(pos[r],1);

            cost[l][r] = inv + take_to_end;
        }
    }

    vector<ll> dp(n+5,inf2);
    dp[0] = 0;

    rep1(i,n){
        rev(j,i,1){
            amin(dp[i], dp[j-1] + cost[j][i]);
        }
    }

    ll ans = dp[n];
    cout << ans << endl;

    /*

    ll n; cin >> n;
    vector<ll> a(n);
    rep(i,n) a[i] = i + 1;

    do{

        bool good = true;

        rep(i,n-1){
            if(a[i] < a[i+1] + 2){
                conts;
            }

            good = false;
            break;
        }

        if(good){
            debug(a);
        }

    } while(next_permutation(all(a)));

    */
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

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