제출 #1269477

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1269477		tin_le	Longest beautiful sequence (IZhO17_subsequence)	C++20	100 / 100	4463 ms	174380 KiB

subsequence

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define vt vector
#define all(x) begin(x), end(x)
#define allr(x) rbegin(x), rend(x)
#define ub upper_bound
#define lb lower_bound
#define db double
#define ld long db
#define ll long long
#define ull unsigned long long
#define vi vt<int>
#define vvi vt<vi>
#define vvvi vt<vvi>
#define pii pair<int, int>
#define vpii vt<pii>
#define vvpii vt<vpii>
#define vll vt<ll>  
#define vvll vt<vll>
#define pll pair<ll, ll>    
#define vpll vt<pll>
#define vvpll vt<vpll>
#define ar(x) array<int, x>
#define var(x) vt<ar(x)>
#define vvar(x) vt<var(x)>
#define al(x) array<ll, x>
#define vall(x) vt<al(x)>
#define vvall(x) vt<vall(x)>
#define vs vt<string>
#define pb push_back
#define ff first
#define ss second
#define rsz resize
#define sum(x) (ll)accumulate(all(x), 0LL)
#define srt(x) sort(all(x))
#define srtR(x) sort(allr(x))
#define srtU(x) sort(all(x)), (x).erase(unique(all(x)), (x).end())
#define rev(x) reverse(all(x))
#define MAX(a) *max_element(all(a)) 
#define MIN(a) *min_element(all(a))
#define SORTED(x) is_sorted(all(x))
#define ROTATE(a, p) rotate(begin(a), begin(a) + p, end(a))
#define i128 __int128
#define IOS ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#if defined(LOCAL) && __has_include("debug.h")
  #include "debug.h"
#else
  #define debug(...)
  #define startClock
  #define endClock
  inline void printMemoryUsage() {}
#endif
template<class T> using max_heap = priority_queue<T>; template<class T> using min_heap = priority_queue<T, vector<T>, greater<T>>;
template<typename T, size_t N> istream& operator>>(istream& is, array<T, N>& arr) { for (size_t i = 0; i < N; i++) { is >> arr[i]; } return is; }
template<typename T, size_t N> istream& operator>>(istream& is, vector<array<T, N>>& vec) { for (auto &arr : vec) { is >> arr; } return is; }
template<typename T1, typename T2>  istream &operator>>(istream& in, pair<T1, T2>& input) { return in >> input.ff >> input.ss; }
template<typename T> istream &operator>>(istream &in, vector<T> &v) { for (auto &el : v) in >> el; return in; }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const static ll INF = 1e18;
const static int inf = 1e9 + 100;
const static int MX = 1e5 + 5;

const int K = 20;
const int L = 10;
pii a[1 << L][1 << L][K + 1];
void update(int mask, pii v) {
    int h = (mask >> L);
    int l = mask & ((1 << L) - 1);
    for(int i = 0; i < 1 << L; i++) {
        int pc = __builtin_popcount(i & h);
        a[l][i][pc] = max(a[l][i][pc], v);
    }
}

pii query(int mask, int k) {
    int h = (mask >> L);
    int l = mask & ((1 << L) - 1);
    pii ans = {};
    for(int i = 0; i < 1 << L; i++) {
        int p = __builtin_popcount(i & l);
        if(p <= k) {
            ans = max(ans, a[i][h][k - p]); 
        }
    }
    return ans;
}

void solve() {
    int n; cin >> n;
    vi A(n + 1), B(n + 1);
    for(int i = 1; i <= n; i++) cin >> A[i];
    for(int i = 1; i <= n; i++) cin >> B[i];
    vi par(n + 1), dp(n + 1);
    int best = 0;
    for(int i = 1; i <= n; i++) {
        auto curr = query(A[i], B[i]);
        par[i] = curr.ss;
        curr.ff += 1;
        dp[i] = curr.ff;
        if(dp[i] > dp[best]) best = i;
        curr.ss = i;
        update(A[i], curr);
    }
    vi ans;
    while(best) {
        ans.pb(best);
        best = par[best];
    }
    rev(ans);
    cout << ans.size() << '\n';
    for(auto& x : ans) cout << x << ' ';
    cout << '\n';
}

signed main() {
    IOS;
    startClock
    int t = 1;
    //cin >> t;
    for(int i = 1; i <= t; i++) {   
        //cout << "Case #" << i << ": ";  
        solve();
    }
    endClock;
    printMemoryUsage();
    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 16 / 16
• Subtask 3: 17 / 17
• Subtask 4: 60 / 60