제출 #1261670

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1261670		biank	Teams (IOI15_teams)	C++20	100 / 100	914 ms	183592 KiB

teams

#include "teams.h"
#include <bits/stdc++.h>
 
#define forsn(i, s, n) for (int i = int(s); i < int(n); i++)
#define forn(i, n) forsn(i, 0, n)
#define dforsn(i, s, n) for (int i = int(n) - 1; i >= int(s); i--)
#define dforn(i, n) dforsn(i, 0, n)
 
#define sz(x) int(x.size())
#define all(x) begin(x), end(x)
 
#define pb push_back
#define eb emplace_back
 
#define fst first
#define snd second
 
using namespace std;
 
using vi = vector<int>;
using ii = pair<int, int>;

const int SZ = 1 << 19;

vi st(1, 0), l(1, 0), r(1, 0);

int newNode(int u) {
    st.pb(st[u]), l.pb(l[u]), r.pb(r[u]);
    return sz(st) - 1;
}

int update(int node, int pos, int val, int lo = 0, int hi = SZ) {
    node = newNode(node);
    if (hi - lo == 1) {
        st[node] += val;
        return node;
    }
    int mid = (lo + hi) / 2;
    if (pos < mid) {
        l[node] = update(l[node], pos, val, lo, mid);
    } else {
        r[node] = update(r[node], pos, val, mid, hi);
    }
    st[node] = st[l[node]] + st[r[node]];
    return node;
}

int query(int s, int e, int node1, int node2, int lo, int hi) {
    if (e <= lo || hi <= s) return 0;
    if (s <= lo && hi <= e) return st[node2] - st[node1];
    int mid = (lo + hi) / 2;
    return query(s, e, l[node1], l[node2], lo, mid) + query(s, e, r[node1], r[node2], mid, hi);
}

int find(int need, int node1, int node2, int lo, int hi) {
    if (hi - lo == 1) return lo;
    int mid = (lo + hi) / 2;
    int leftSum = st[l[node2]] - st[l[node1]];
    if (leftSum >= need) {
        return find(need, l[node1], l[node2], lo, mid);
    }
    return find(need - leftSum, r[node1], r[node2], mid, hi);
}

int n;
int roots[SZ];

int query(int l1, int r1, int l2, int r2) {
    return query(l1, r1, roots[l2 - 1], roots[r2 - 1], 0, SZ);
}

int find(int k, int l, int r) {
    return find(k, roots[l - 1], roots[r - 1], 0, SZ);
}

void init(int N, int A[], int B[]) {
    n = N;
    st.reserve(1e7), l.reserve(1e7), r.reserve(1e7);
    vector<vi> byA(n + 1);
    forn(i, n) byA[A[i]].pb(B[i]);
    roots[0] = 0;
    forsn(i, 1, n + 1) {
        roots[i] = roots[i - 1];
        for (int b : byA[i]) roots[i] = update(roots[i], b, 1);
    }
}

int can(int M, int K[]) {
    vi a, b, cnt;
    sort(K, K + M);
    a.pb(0), b.pb(n), cnt.pb(0);
    forn(i, M) {
        while (!b.empty() && b.back() < K[i]) {
            a.pop_back(), b.pop_back(), cnt.pop_back();
        }
        int prevB = K[i], need = K[i];
        while (!cnt.empty()) {
            int total = query(prevB, b.back() + 1, a.back() + 1, K[i] + 1) + cnt.back();
            if (total < need) {
                need -= total, prevB = b.back() + 1;
                a.pop_back(), b.pop_back(), cnt.pop_back();
            } else {
                break;
            }
        }
        if (cnt.empty()) return 0;
        int prev = query(0, prevB, a.back() + 1, K[i] + 1);
        int minB = find(prev + need, a.back() + 1, K[i] + 1);
        minB = min(minB, b.back());
        int total = query(prevB, minB + 1, a.back() + 1, K[i] + 1);
        if (minB == b.back()) total += cnt.back(), cnt.back() = 0;
        assert(total >= need);
        a.pb(K[i]), b.pb(minB), cnt.pb(total - need); 
    }
    return 1;
}

• Subtask 1: 21 / 21
• Subtask 2: 13 / 13
• Subtask 3: 43 / 43
• Subtask 4: 23 / 23