Submission #578766

#	Time	Username	Problem	Language	Result	Execution time	Memory
578766		jeroenodb	Longest beautiful sequence (IZhO17_subsequence)	C++17	40 / 100	6101 ms	175132 KiB

subsequence

#include "bits/stdc++.h"
using namespace std;
#define all(x) begin(x),end(x)
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { string sep; for (const T &x : v) os << sep << x, sep = " "; return os; }
#define debug(a) cerr << "(" << #a << ": " << a << ")\n";
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pi;
const int mxN = 1e5+1, oo = 1e9;
const int B = 10;
struct DS {
    pi best[1<<B][1<<B][B*2+1] = {};

    static void cmax(pi& a, const pi& b) {
        a = max(a,b);
    }
    DS() {}
    static pi split(int a) {
        return {a&((1<<B)-1),a>>B};
    }
    void insert(int a, pi v) {
        auto [l,r] = split(a);
        for(int msk=0;msk<1<<B;++msk) {
            cmax(best[msk][r][__builtin_popcount(msk&a)],v);
        }
    }
    pi query(int a, int k) {
        auto [l,r] = split(a);
        pi res = {};
        for(int msk=0;msk<1<<B;++msk) {
            int pp = __builtin_popcount(msk&r);
            if(pp<=k) {
                cmax(res,best[l][msk][k-pp]);
            }
        }
        return res;
    }
} *ds = new DS();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n; cin >> n;
    vi a(n);
    vector<short> k(n);
    for(auto& i : a) cin >> i;
    for(auto& i : k) cin >> i;
    vi dp(n),par(n,-1);
    for(int i=0;i<n;++i) {
        auto [v,pr] = ds->query(a[i],k[i]);
        if(v) par[i]=pr;
        dp[i]=v+1;
        ds->insert(a[i],{dp[i],i});
    }
    // recover solution
    auto it = max_element(all(dp));
    int i = it-dp.begin();
    cout << *it << '\n';
    vi b;
    while(i!=-1) {
        b.push_back(i+1);
        i=par[i];
    }
    reverse(all(b));
    cout << b << '\n';
}

• Subtask 1: 7 / 7
• Subtask 2: 16 / 16
• Subtask 3: 17 / 17
• Subtask 4: 0 / 60