Submission #1264966

#	Time	Username	Problem	Language	Result	Execution time	Memory
1264966		btninh	Harbingers (CEOI09_harbingers)	C++20	0 / 100	107 ms	29336 KiB

harbingers

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

struct Line {
    mutable ll k, m, p;
    bool operator<(const Line& o) const { return k < o.k; }
    bool operator<(ll x) const { return p < x; }
};

struct LineContainer : multiset<Line, less<>> {
    static const ll inf = LLONG_MAX;

    ll divFloor(ll a, ll b) {
        return a / b - ((a ^ b) < 0 && a % b);
    }

    bool isect(iterator x, iterator y) {
        if (y == end()) { x->p = inf; return false; }
        if (x->k == y->k) x->p = (x->m > y->m ? inf : -inf);
        else x->p = divFloor(y->m - x->m, x->k - y->k);
        return x->p >= y->p;
    }

    struct Change { int type; Line val; };
    // type=0: previously ERASED (we should re-insert on rollback)
    // type=1: previously INSERTED (we should remove on rollback)

    vector<Change> history;

    // helper: find an element in multiset with matching (k,m)
    iterator find_km(ll k, ll m) {
        // find lower_bound by key k
        auto it = lower_bound({k, m, 0});
        // scan rightwards through equal k's to find matching m
        while (it != end() && it->k == k) {
            if (it->m == m) return it;
            ++it;
        }
        // also scan leftwards a bit in case lower_bound returned later equal slot
        it = lower_bound({k, m, 0});
        if (it != begin()) {
            auto jt = it;
            do {
                --jt;
                if (jt->k != k) break;
                if (jt->m == m) return jt;
                if (jt == begin()) break;
            } while (true);
        }
        return end();
    }

    iterator add(ll k, ll m) {
        auto z = insert({k,m,0}), y = z++, x = y;
        history.push_back({1,*y}); // record insertion

        while (isect(y,z)) {
            history.push_back({0,*z});
            z = erase(z);
        }
        if (x != begin() && isect(--x,y)) {
            history.push_back({0,*y});
            erase(y);
        }
        while ((y=x) != begin() && (--x)->p >= y->p) {
            history.push_back({0,*y});
            erase(y);
        }
        // return iterator to inserted line (may be invalid if erased later, but we use history to rollback)
        // find current position of inserted (k,m)
        auto itcur = find_km(k,m);
        return itcur;
    }

    void rollback(int snap) {
        while ((int)history.size() > snap) {
            auto ch = history.back(); history.pop_back();
            if (ch.type == 0) {
                // re-insert the erased line
                insert(ch.val);
            } else {
                // remove one inserted line matching (k,m)
                auto it = find_km(ch.val.k, ch.val.m);
                if (it != end()) erase(it);
                // else: should not happen, but ignore safely
            }
        }
    }

    int snapshot() { return history.size(); }

    ll query(ll x) {
        assert(!empty());
        auto l = *lower_bound(x);
        return l.k * x + l.m;
    }
};


int N;
vector<vector<pair<int,ll>>> adj;
vector<ll> S, V, dist, opt;
LineContainer cht;

void dfs(int u, int p) {
    if (u == 1) {
        opt[u] = 0;
    } else {
        // query: code uses sign-flip trick -> stored lines are (k = +dist, m = -opt)
        ll best_neg = cht.query(V[u]); // this returns max of (-opt[j] + dist[j]*V[u])
        ll best = -best_neg;
        opt[u] = S[u] + V[u]*dist[u] + best;
    }

    int snap = cht.snapshot();
    // add line for u: we store (k = +dist[u], m = -opt[u]) so hull is for max
    cht.add(dist[u], -opt[u]);

    for (auto [v,w] : adj[u]) if (v != p) {
        dist[v] = dist[u] + w;
        dfs(v,u);
    }

    cht.rollback(snap);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    if (!(cin >> N)) return 0;
    adj.assign(N+1, {});
    for (int i=0;i<N-1;i++){
        int u,v; ll w; cin >> u >> v >> w;
        adj[u].push_back({v,w});
        adj[v].push_back({u,w});
    }
    S.assign(N+1,0);
    V.assign(N+1,0);
    dist.assign(N+1,0);
    opt.assign(N+1,0);

    for (int i=2;i<=N;i++) cin >> S[i] >> V[i];

    dfs(1,0);

    for (int i=2;i<=N;i++) cout << opt[i] << "\n";
    return 0;
}

• Subtask 1: 0 / 100