Submission #708790

#	Time	Username	Problem	Language	Result	Execution time	Memory
708790		PixelCat	Two Dishes (JOI19_dishes)	C++14	65 / 100	605 ms	58424 KiB

dishes

#include <bits/stdc++.h>
#define For(i, a, b) for(int i = a; i <= b; i++)
#define Forr(i, a, b) for(int i = a; i >= b; i--)
#define F first
#define S second
#define sz(x) ((int)x.size())
#define all(x) x.begin(), x.end()
#define eb emplace_back
#define int LL
using namespace std;
using LL = long long;
using pii = pair<int, int>;

// const int MAXN = 1000010;
const int MAXN = 200010;
const int INF = 8e18;

// range add, range max
#define L(id) ((id) * 2 + 1)
#define R(id) ((id) * 2 + 2)
struct SegTree {
    struct Node {
        int mx, add;
    } a[MAXN << 2];
    void init(int val) {
        For(i, 0, (MAXN << 2) - 1) {
            a[i].mx = val;
            a[i].add = 0;
        }
    }
    void _add(int id, int x) {
        a[id].add += x;
        a[id].mx += x;
    }
    void pull(int id) {
        a[id].mx = max(a[L(id)].mx, a[R(id)].mx);
    }
    void push(int id) {
        if(a[id].add != 0) {
            _add(L(id), a[id].add);
            _add(R(id), a[id].add);
            a[id].add = 0;
        }
    }
    void add(int id, int l, int r, int L, int R, int x) {
        // if(l > R || r < L) return;
        if(l >= L && r <= R) {
            _add(id, x);
            return;
        }
        push(id);
        int m = (l + r) / 2;
        if(L <= m) add(L(id), l, m, L, R, x);
        if(R > m)  add(R(id), m + 1, r, L, R, x);
        pull(id);
    }
    void set(int id, int l, int r, int i, int x) {
        if(l == r) {
            a[id].mx = x;
            a[id].add = 0;
            return;
        }
        push(id);
        int m = (l + r) / 2;
        if(i <= m) set(L(id), l, m, i, x);
        else       set(R(id), m + 1, r, i, x);
        pull(id);
    }
    int ask(int id, int l, int r, int L, int R) {
        if(l >= L && r <= R) return a[id].mx;
        push(id);
        int m = (l + r) / 2;
        int res = -INF;
        if(L <= m) res = max(res, ask(L(id), l, m, L, R));
        if(R > m)  res = max(res, ask(R(id), m + 1, r, L, R));
        return res;
    }
} seg, seg2, seg3;

int a1[MAXN];
int s1[MAXN];
int p1[MAXN];
int pr1[MAXN];
int t1[MAXN];

int a2[MAXN];
int s2[MAXN];
int p2[MAXN];
int pr2[MAXN];
int t2[MAXN];

// int su[MAXN][MAXN];
int dp[MAXN];

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    // NYA =^-w-^=
    int n, m;
// #define QWQ
#ifndef QWQ
    cin >> n >> m;
    assert(n <= MAXN);
    For(i, 1, n) {
        cin >> a1[i] >> s1[i] >> p1[i];
        pr1[i] = pr1[i - 1] + a1[i];
    }
    For(i, 1, m) {
        cin >> a2[i] >> s2[i] >> p2[i];
        pr2[i] = pr2[i - 1] + a2[i];
    }
#else
    mt19937 rng(48763);
    n = m = 5;
    For(i, 1, n) {
        a1[i] = rng() % 10 + 1;
        s1[i] = rng() % 100 + 1;
        p1[i] = 1;
        pr1[i] = pr1[i - 1] + a1[i];
    }
    For(i, 1, m) {
        a2[i] = rng() % 10 + 1;
        s2[i] = rng() % 100 + 1;
        p2[i] = 1;
        pr2[i] = pr2[i - 1] + a2[i];
    }
#endif

    vector<pii> v;
    For(i, 1, n) {
        auto it = upper_bound(pr2, pr2 + m + 1, s1[i] - pr1[i]);
        t1[i] = it - pr2 - 1;
        if(t1[i] >= 0) v.eb(t1[i], i);
    }
    sort(all(v));
    reverse(all(v));
    
    For(i, 1, m) {
        auto it = upper_bound(pr1, pr1 + n + 1, s2[i] - pr2[i]);
        t2[i] = it - pr1 - 1;
    }
    
    seg.init(-INF);
    seg2.init(0);
    for(auto &i:v) {
        seg2.add(0, 0, n, i.S, n, p1[i.S]);
    }
    seg.set(0, 0, n, 0, 0);
    
    For(it, 0, m) {
        if(it && (t2[it] >= 0)) {
            seg.add(0, 0, n, 0, t2[it], p2[it]);
        }
        while(sz(v) && v.back().F == it) {
            int i = v.back().S;
            v.pop_back();

            seg2.add(0, 0, n, i, n, -p1[i]);
            seg.add(0, 0, n, i, n, p1[i]);
            
            int val = seg.ask(0, 0, n, 0, i - 1) + seg2.ask(0, 0, n, i, i) + p1[i];
            seg.set(0, 0, n, i, val - seg2.ask(0, 0, n, i, i));
        }
    }
    cout << seg.ask(0, 0, n, 0, n) << "\n";
    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 3 / 3
• Subtask 3: 7 / 7
• Subtask 4: 39 / 39
• Subtask 5: 11 / 11
• Subtask 6: 0 / 9
• Subtask 7: 0 / 17
• Subtask 8: 0 / 9