Submission #423156

#	Time	Username	Problem	Language	Result	Execution time	Memory
423156		abdzag	Financial Report (JOI21_financial)	C++17	100 / 100	1101 ms	53700 KiB

Main

#include <bits/stdc++.h>

using namespace std;

#define FAST_IO ios_base::sync_with_stdio(0); cin.tie(nullptr)
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(n) FOR(O, 1, (n))
#define f first
#define s second
#define pb push_back
typedef pair<int, int> pii;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<pii> vii;
typedef vector<ll> vl;

const int MAXN = 300100, MAXK = 20;
const ll INF = 1e17;

// DSU
int par[MAXN], mn[MAXN];
int find(int a) { return par[a] = par[a] == a ? a : find(par[a]); }
bool same(int a, int b) { return find(a) == find(b); }
void unite(int a, int b) {
    a = find(a), b = find(b);
    if (a == b) return;
    par[b] = a;
    mn[a] = min(mn[a], mn[b]);
}

int n, d;
int a[MAXN];
set<int> tmp;
unordered_map<int, int> toId;

// two segtrees. i am actually using only one in this solution (the second one was left from the solution for subtasks)
int tree[2][4 * MAXN];

void upd(int tid, int p, int val, int id = 1, int le = 1, int ri = n) {
    if (le == ri) {
        tree[tid][id] = val;
        return;
    }
    int mid = (le + ri) / 2;
    if (p <= mid)
        upd(tid, p, val, 2 * id, le, mid);
    else
        upd(tid, p, val, 2 * id + 1, mid + 1, ri);
    tree[tid][id] = max(tree[tid][2 * id], tree[tid][2 * id + 1]);
}

int get(int tid, int x, int y, int id = 1, int le = 1, int ri = n) {
    if (x > ri || y < le) return 0;
    if (x <= le && ri <= y) return tree[tid][id];
    int mid = (le + ri) / 2;
    return max(get(tid, x, y, 2 * id, le, mid), get(tid, x, y, 2 * id + 1, mid + 1, ri));
}

bool cmp(int f, int s) {
    if (a[f] != a[s])
        return a[f] < a[s];
    return f > s;
}

vi seq;
set<int> spec;

int main()
{
    FAST_IO;



    cin >> n >> d;
    FOR(i, 1, n) cin >> a[i];
    FOR(i, 1, n) par[i] = i, mn[i] = i;

    // coordinate compression which ended up being not necessary in the full solution
    FOR(i, 1, n) tmp.insert(a[i]);
    int cur = 0;
    for (int x : tmp)
        toId[x] = ++cur;
    FOR(i, 1, n) a[i] = toId[a[i]];

    FOR(i, 1, n) upd(0, i, a[i]);

    FOR(i, 1, n) seq.pb(i);

    // sorting by a[i]
    sort(seq.begin(), seq.end(), cmp);

    for (int i : seq) {

        auto it = spec.lower_bound(i);
        if (it != spec.end()) {
            int id = *it;
            if (abs(id - i) <= d) {
                unite(id, i);
            }
        }
        if (it != spec.begin()) {
            it--;
            int id = *it;
            if (abs(id - i) <= d) {
                unite(id, i);
            }
        }

        spec.insert(i);

        int cur = 1; // cur is the value of dp[i]
        int fr = mn[find(i)];
        if (fr <= i - 1) cur = max(cur, get(1, fr, i - 1) + 1);
        upd(1, i, cur);
    }

    cout << (get(1, 1, n)) << "\n";

    return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 20 / 20
• Subtask 4: 12 / 12
• Subtask 5: 5 / 5
• Subtask 6: 35 / 35