Submission #465551

#	Time	Username	Problem	Language	Result	Execution time	Memory
465551		Sench2006	Exam (eJOI20_exam)	C++14	23 / 100	165 ms	8180 KiB

exam

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;

#define ll long long
#define ar array

#define f first
#define s second
#define mpr make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound

#define FOR(i, a, b) for(auto i = a; i < b; i++)
#define FORD(i, a, b) for(auto i = a - 1; i >= b; i--)
#define trav(x, v) for(auto x : v)

#define fastio ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fileio freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);

// --------------- SOLUTION --------------- //

const int N = 1e5+5;
int n;
int a[N], b[N];
int t[4 * N];
int next_max[N], prev_max[N];
int dp[5001][5001];
map <int, int> mp;

void pre() {
    stack<int> s;
    FOR(i, 0, n) {
        while (!s.empty() && a[s.top()] <= a[i])
            s.pop();
        if (s.empty())
            next_max[i] = -1;
        else
            next_max[i] = s.top();
        s.push(i);
    }
    while (!s.empty())
        s.pop();
    FORD(i, n, 0) {
        while (!s.empty() && a[s.top()] <= a[i])
            s.pop();
        if (s.empty())
            prev_max[i] = n + 1;
        else
            prev_max[i] = s.top();
        s.push(i);
    }
}

void ubd(int tl, int tr, int x, int y, int pos) {
    if (tl == tr) {
        t[pos] = y;
        return;
    }
    int m = (tl + tr) / 2;
    if (x <= m)
        ubd(tl, m, x, y, pos * 2);
    else
        ubd(m + 1, tr, x, y, pos * 2 + 1);
    t[pos] = max(t[pos * 2], t[pos * 2 + 1]);
}

int qry(int tl, int tr, int l, int r, int pos) {
    if (l > r)
        return 0;
    if (tl == l && tr == r)
        return t[pos];
    int m = (tl + tr) / 2;
    return max(qry(tl, m, l, min(m, r), pos * 2),
               qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));

}

int segTreeSolve() {
    FOR(i, 0, n) {
        mp[a[i]] = i;
    }
    FOR(i, 0, n) {
        if (mp.find(b[i]) == mp.end()) continue;
        if (next_max[mp[b[i]]] < i && prev_max[mp[b[i]]] > i) ubd(0, n - 1, mp[b[i]], qry(0, n - 1, 0, mp[b[i]], 1) + 1, 1);
    }
    return t[1];
}

int less5000() {
    int ans = 0;
    FOR(i, 0, n) {
        FOR(j, 0, n) {
            if (a[i] == b[j] && next_max[i] < j && j < prev_max[i]) {
                dp[i][j + 1] = max(dp[i][j + 1], dp[i][j] + 1);
            }
            dp[i][j + 1] = max(dp[i][j + 1], dp[i][j]);
            dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
        }
    }
    FOR(i, 0, n)
        FOR(j, 0, n)
            ans = max(ans, dp[i][j]);
    return ans;
}

void solve() { 
    cin >> n;
    FOR(i, 0, n) {
        cin >> a[i];
    }
    FOR(i, 0, n) {
        cin >> b[i];
    }
    if (n <= 5000) {
        cout << less5000() << "\n";
        return;
    }
    pre();
    cout << segTreeSolve() << "\n";
}

int main() {
    fastio;
    // fileio;
    int tests = 1;
    // cin >> tests;
    while (tests--) {
        solve();
    }

    return 0;
}

• Subtask 1: 0 / 14
• Subtask 2: 0 / 12
• Subtask 3: 0 / 13
• Subtask 4: 23 / 23
• Subtask 5: 0 / 16
• Subtask 6: 0 / 22