oj.uz

답안 #560342

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
560342 2022-05-11T10:10:28 Z kartel 사탕 분배 (IOI21_candies) C++17
38 / 100
 5000 ms 57548 KB 
#include <bits/stdc++.h>

//#include "grader.cpp"
#include "candies.h"

#define F first
#define S second
#define pb push_back
#define sz(x) (int)x.size()
#define el "\n"
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;

struct st_beats {
    struct node {
        ll mx, mn;
        ll D, Dx;

        node() {
            mx = mn = D = 0;
            Dx = -2e9;
        }

        node operator +(node other) {
            node ret;

            ret.mx = max(mx, other.mx);
            ret.mn = min(mn, other.mn);
            return ret;
        }

        void mrg(node &a, node &b) {
            node x = a + b;
            mx = x.mx; mn = x.mn;
        }

        void add() {
            mx += D; mn += D;
        }

        void updL() {
            mx = mn = Dx;
            Dx = -2e9;
        }
    };

    vector <node> t;
    int n, h;

    st_beats() {}

    void init(int _n, int _h) {
        t.resize(8 * _n);
        for (int i = 0; i < sz(t); i++) {
            t[i] = node();
        }
        h = _h;
        n = _n;
    }

    void push(int v) {
        if (t[v].D) {
            t[v].add();
            t[v * 2].D += t[v].D;
            t[v * 2 + 1].D += t[v].D;
            t[v].D = 0;
        }
    }

    void pushL(int v) {
        if (t[v].Dx != -2e9) {
            t[v * 2].Dx = t[v].Dx;
            t[v * 2 + 1].Dx = t[v].Dx;
            t[v * 2].D = t[v * 2 + 1].D = 0;

            t[v].updL();
        }
    }

    void psh(int v) {
        pushL(v);
        push(v);
    }

    void upd(int v, int l, int r, int tl, int tr, int val) {
        psh(v);
        if (l > r || tl > tr || tl > r || l > tr) {
            return;
        }
        int md = (l + r) >> 1;
        if (l >= tl && r <= tr) {
            t[v].D += val;
            psh(v);
            return;
        }
        upd(v * 2, l, md, tl, tr, val);
        upd(v * 2 + 1, md + 1, r, tl, tr, val);
        t[v].mrg(t[v * 2], t[v * 2 + 1]);
    }

    void upd(int v, int l, int r, int x) {
        psh(v);

        if (t[v].mn > h) {
            t[v].Dx = h;
            psh(v);
            return;
        }
        if (t[v].mx < 0) {
            t[v].Dx = 0;
            psh(v);
            return;
        }

        if (l == r || (t[v].mn >= 0 && t[v].mx <= h)) {
            return;
        }

        int md = (l + r) >> 1;
        upd(v * 2, l, md, 0);
        upd(v * 2 + 1, md + 1, r, 0);
        t[v].mrg(t[v * 2], t[v * 2 + 1]);
    }

    void upd(int l, int r, int val) {
        upd(1, 0, n - 1, l, r, val);

        upd(1, 0, n - 1, 0);
    }

    int get(int v, int l, int r, int ps) {
        psh(v);
        if (l == r) {
            return t[v].mx;
        } else {
            int md = (l + r) >> 1;
            if (ps <= md) {
                return get(v * 2, l, md, ps);
            }
            return get(v * 2 + 1, md + 1, r, ps);
        }
    }

    int get(int x) {
        return get(1, 0, n - 1, x);
    }
} stb;

struct st_min {
    vector <ll> t;
    int n;

    st_min() {}

    void build(int v, int l, int r, vector <ll> &a) {
        if (l == r) {
            t[v] = abs(a[l]);
            return;
        } else {
            int md = (l + r) >> 1;
            build(v * 2, l, md, a);
            build(v * 2 + 1, md + 1, r, a);
            t[v] = min(t[v * 2], t[v * 2 + 1]);
        }
    }

    void init(vector <ll> &a) {
        n = sz(a);
        t.resize(8 * n);
        build(1, 0, n - 1, a);
    }

    void upd(int v, int l, int r, int ps, ll val) {
        if (l == r) {
            t[v] = val;
            return;
        } else {
            int md = (l + r) >> 1;
            if (ps <= md) {
                upd(v * 2, l, md, ps, val);
            } else {
                upd(v * 2 + 1, md + 1, r, ps, val);
            }
            t[v] = min(t[v * 2], t[v * 2 + 1]);
        }
    }

    int get(int v, int l, int r, ll c) {
        if (t[v] > c) {
            return -1;
        }
        if (l == r) {
            return l;
        }
        int md = (l + r) >> 1;
        int cur = get(v * 2, l, md, c);
        if (cur == -1) {
            return get(v * 2 + 1, md + 1, r, c);
        }
        return cur;
    }

    void upd(int ps, ll val) {
        upd(1, 0, n - 1, ps, val);
    }

    int get(ll c) {
        return get(1, 0, n - 1, c);
    }
};

vector <int> solve_st_beats(vector <int> &c, vector <int> &l, vector <int> &r, vector <int> &v) {
    int n = sz(c), q = sz(l);

    stb.init(n, c[0]);
    for (int i = 0; i < q; i++) {
        stb.upd(l[i], r[i], v[i]);
    }

    vector <int> a(n);
    for (int i = 0; i < n; i++) {
        a[i] = stb.get(i);
    }

    return a;
}

vector <int> solve_offline(vector <int> &c, vector <int> &l, vector <int> &r, vector <int> &v) {
    vector <ll> p = {v[0]};
    int q = sz(l), n = sz(c);
    vector <int> nxt(q, 0);
    set <pair <int, ll> > se;

    auto sg = [&](ll x) {
        if (x < 0) {
            return -1;
        }
        return 1;
    };

    for (int i = 1; i < q; i++) {
        if (sg(p.back()) == sg(v[i])) {
            p.back() += v[i];
        } else {
            p.pb(v[i]);
        }
    }
    if (p[0] < 0) {
        p.erase(p.begin());
    }
    q = sz(p);
    for (int i = 0; i < q; i++) {
        nxt[i] = i + 1;
        se.insert({i, p[i]});
    }

    vector <int> ind(n, 0);
    iota(all(ind), 0);
    sort(all(ind), [&](int i, int j) {return (c[i] < c[j]);});

    st_min tM;
    tM.init(p);

    vector <int> ans(n, 0);
    for (auto i : ind) {
        ll x = c[i];
        int j = tM.get(x);
        while (j != -1) {
            ll new_val = p[j];
            int was = j;
            se.erase({j, p[j]});
            vector <int> pos;
            while (nxt[j] < q && new_val >= 0 && new_val <= x) {
                j = nxt[j];
                pos.pb(j);
                se.erase({j, p[j]});
                new_val += p[j];
            }
            for (auto x : pos) {
                nxt[x] = j;
                tM.upd(x, 2e18);
            }
            if (was != j) {
                nxt[was] = j;
            }
            se.insert({was, new_val});
            tM.upd(was, abs(new_val));
            if (nxt[j] == q) {
                break;
            }
            j = tM.get(x);
        }
        ll val = se.rbegin() -> S;
        if (val < 0) {
            ans[i] = max(0ll, x + val);
        } else {
            ans[i] = min(x, val);
        }
    }

    return ans;
}

vector <int> distribute_candies(vector <int> c, vector <int> l, vector <int> r, vector <int> v) {
    int n = sz(c), q = sz(l);

    if (n <= 2000 && q <= 2000) {
        vector <int> a(n, 0);

        for (int i = 0; i < q; i++) {
            for (int j = l[i]; j <= r[i]; j++) {
                a[j] += v[i];

                a[j] = max(a[j], 0);
                a[j] = min(a[j], c[j]);
            }
        }
        return a;
    } else {
        bool g = 1;
        for (int i = 0; i < q; i++) {
            g &= (v[i] > 0);
        }

        if (g) {
            vector <ll> pf(n, 0);
            vector <int> a(n, 0);

            for (int i = 0; i < q; i++) {
                pf[l[i]] += v[i];
                if (r[i] + 1 < n) {
                    pf[r[i] + 1] -= v[i];
                }
            }

            for (int i = 1; i < n; i++) {
                pf[i] += pf[i - 1];
            }
            for (int i = 0; i < n; i++) {
                a[i] = min(max(0ll, pf[i]), (ll)c[i]);
            }

            return a;
        } else {
            bool eq_c = 1;

            for (int i = 1; i < n; i++) {
                eq_c &= (c[i] == c[0]);
            }

            if (eq_c) {
                return solve_st_beats(c, l, r, v);
            }
            return solve_offline(c, l, r, v);
        }
    }
}

/*
3
10 15 13

2
0 2  20
0 2 -11

*/

Subtask #1
3.0 / 3.0

# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 0 ms 212 KB Output is correct
3 Correct 1 ms 320 KB Output is correct
4 Correct 1 ms 340 KB Output is correct
5 Correct 4 ms 340 KB Output is correct

Subtask #2
8.0 / 8.0

# 결과 실행 시간 메모리 Grader output
1 Correct 105 ms 8812 KB Output is correct
2 Correct 96 ms 8804 KB Output is correct
3 Correct 108 ms 8896 KB Output is correct

Subtask #3
27.0 / 27.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 212 KB Output is correct
2 Correct 158 ms 5516 KB Output is correct
3 Correct 93 ms 53928 KB Output is correct
4 Correct 364 ms 57448 KB Output is correct
5 Correct 413 ms 57460 KB Output is correct
6 Correct 580 ms 57436 KB Output is correct
7 Correct 450 ms 57548 KB Output is correct

Subtask #4
0 / 29.0

# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Execution timed out 5090 ms 19044 KB Time limit exceeded
4 Halted 0 ms 0 KB -

Subtask #5
0 / 33.0

# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 0 ms 212 KB Output is correct
3 Correct 1 ms 320 KB Output is correct
4 Correct 1 ms 340 KB Output is correct
5 Correct 4 ms 340 KB Output is correct
6 Correct 105 ms 8812 KB Output is correct
7 Correct 96 ms 8804 KB Output is correct
8 Correct 108 ms 8896 KB Output is correct
9 Correct 1 ms 212 KB Output is correct
10 Correct 158 ms 5516 KB Output is correct
11 Correct 93 ms 53928 KB Output is correct
12 Correct 364 ms 57448 KB Output is correct
13 Correct 413 ms 57460 KB Output is correct
14 Correct 580 ms 57436 KB Output is correct
15 Correct 450 ms 57548 KB Output is correct
16 Correct 0 ms 212 KB Output is correct
17 Correct 1 ms 212 KB Output is correct
18 Execution timed out 5090 ms 19044 KB Time limit exceeded
19 Halted 0 ms 0 KB -