제출 #571748

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
571748		Vanilla	Financial Report (JOI21_financial)	C++17	100 / 100	504 ms	27380 KiB

Main

#include <bits/stdc++.h>
#include <fstream>
#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>;
struct custom_hash {static uint64_t splitmix64(uint64_t x) {x += 0x9e3779b97f4a7c15;x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;x = (x ^ (x >> 27)) * 0x94d049bb133111eb;return x ^ (x >> 31);}size_t operator()(uint64_t x) const {static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();return splitmix64(x + FIXED_RANDOM);}};
// string __fname = ""; ifstream in (__fname + ".in"); ofstream out (__fname + ".out"); 
// #define cin in 
// #define cout out
#define int64 long long
#define uint64 unsigned long long
#define x first
#define y second
#define pb push_back
#define pii pair <int, int> 
#define pii64 pair <int64, int64>
#define db(x) cout << "> " << #x << ": " << (x) << "\n"
#define qr queries()
#define yn(x) if (x) {ctn("Yes");}else {ctn("No");}
void solve(int);
void queries(){int n;cin >> n;for (int i = 1; i <= n; i++) solve(i);}
template<class T>T ceildiv(T a, T b) {return a / b + !!(a % b);}
template<class T>T gcd (T a, T b){return (b ? gcd(b, a % b): a);}
template<class T>T lcm (T a, T b){return a * b / gcd(a, b);}
// // // // // // // // // // // // // // // // // // // // // // 
/*                  TEMPLATE - VANILLA                         */
// // // // // // // // // // // // // // // // // // // // // //
const int ddx[] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int ddy[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
const double pi = 3.14159265359;
const double eps = 1e-6;
const int64 hash_inv = 940594066;
const int64 hash_p = 101;
const int64 mod = 1e9 + 7;
const int maxn = 3e5 + 2;
int dp[maxn];
int a[maxn];
int seg [4 * maxn];
int dad [maxn];
int n, d;

int get_dad(int x) {
    return dad[x] = (x == dad[x] ? x: get_dad(dad[x]));
}

void merge (int x, int y) {
    x = get_dad(x);
    y = get_dad(y);
    if (x < y) dad[y] = x;
    else dad[x] = y;
}
 
int query (int nd, int l, int r, int _l, int _r) {
    if (l > _r || r < _l) return 0;
    if (l >= _l && r <= _r) return seg[nd];
    int mid = (l + r) / 2;
    return max(query(2 * nd, l, mid, _l, _r), query(2 * nd + 1, mid + 1, r, _l, _r));
}
 
int update (int nd, int l, int r, int x) {
    if (l > x || r < x) return seg[nd];
    if (l == r) return seg[nd] = dp[x];
    int mid = (l + r) / 2; 
    return seg[nd] = max(update(2 * nd, l, mid, x), update(2 * nd + 1, mid + 1, r, x));
}
 
void solve(int id){
    
    return;
}
 
bool comp (pii a, pii b) {
    if (a.x != b.x) return a.x < b.x;
    return a.y > b.y;
}
 
int main(){
    ios_base::sync_with_stdio(0); cin.tie(0); cout << fixed; cout << setprecision(10); 
    cin >> n >> d;
    vector <pii> id;
    for (int i = 0; i < maxn; i++){
        dad[i] = i;
    }
    for (int i = 0; i < n; i++){
        cin >> a[i];
        id.pb({a[i], i});
    }
    sort(id.begin(), id.end(), comp);
    set <int> s;
    for (int i = 0; i < n; i++){
        int idx = id[i].y;
        auto it = s.lower_bound(idx);
        if (it != s.end()) {
            if (abs(*it - idx) <= d) {
                merge(*it, idx);
            }
        }
        if (it != s.begin()) {
            it--;
            if (abs(*it - idx) <= d) {
                merge(*it, idx);
            }
        }
        s.insert(idx);
        int l = get_dad(idx);
        dp[idx] = query(1, 0, n - 1, max(0, min(idx - d, l)), idx - 1) + 1;
        update(1, 0, n-1, idx);
    }
    cout << *max_element(dp, dp + 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