제출 #471803

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
471803		dutinmeow	Harbingers (CEOI09_harbingers)	C++17	40 / 100	163 ms	65540 KiB

harbingers

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

using ll = long long;
using pll = pair<ll, ll>;

/**
 * A Persistant Li Chao Impelementation
 * Useful for Convex Hull Trick on Trees
 * Complexity: O(log C)
 * Verified: https://judge.yosupo.jp/problem/line_add_get_min
 */ 
struct PersistantLiChao {
    struct Line {
        int64_t m, b;
        Line(int64_t _m = 0, int64_t _b = numeric_limits<int64_t>::min()) : m(_m), b(_b) {}
        int64_t operator()(int64_t x) { return m * x + b; }
        friend ostream& operator<<(ostream &os, const Line &l) { return os << l.m << "x + " << l.b; }
        ~Line() {}
    };

    struct Node {
        Line line;
        size_t left = -1, right = -1;
        Node(size_t _left, size_t _right) : left(_left), right(_right) {}
        Node(Line _line = Line(), size_t _left = -1, size_t _right = -1) : line(_line), left(_left), right(_right) {}
    };

    vector<int> roots;
    vector<Node> nodes;

private:
    int64_t x_min, x_max;

    int node(Line line, int left = -1, int right = -1) { 
        nodes.emplace_back(line, left, right);
        return nodes.size() - 1;
    }

    int add(int root, Line line, int64_t l, int64_t r) {
        if (root == -1)
            return node(line);
        int64_t root_l = nodes[root].line(l), root_r = nodes[root].line(r);
        int64_t line_l = line(l), line_r = line(r);

        if (root_l >= line_l && root_r >= line_r)
            return root;
        if (root_l <= line_l && root_r <= line_r) 
            return node(line, nodes[root].left, nodes[root].right);
        
        int64_t m = (l + r) / 2;
        int64_t root_m = nodes[root].line(m), line_m = line(m);
        if (root_m > line_m) {
            if (line_l >= root_l)
                return node(nodes[root].line, add(nodes[root].left, line, l, m), nodes[root].right);
            else 
                return node(nodes[root].line, nodes[root].left, add(nodes[root].right, line, m + 1, r));
        } else {
            if (root_l >= line_l)
                return node(line, add(nodes[root].left, nodes[root].line, l, m), nodes[root].right);
            else 
                return node(line, nodes[root].left, add(nodes[root].right, nodes[root].line, m + 1, r));
        }
    }

    int64_t query(int root, int64_t x, int64_t l, int64_t r) {
        if (root == -1)
            return numeric_limits<int64_t>::min();
        if (l == r)
            return nodes[root].line(x);
        
        int64_t m = (l + r) / 2;
        if (x <= m)
            return max(nodes[root].line(x), query(nodes[root].left, x, l, m));
        else 
            return max(nodes[root].line(x), query(nodes[root].right, x, m + 1, r));
    }

public:
    PersistantLiChao(int N, int64_t _x_min, int64_t _x_max) : x_min(_x_min), x_max(_x_max) {
        roots.resize(N, -1);
    }

    void add(int r, const int64_t &m, const int64_t &b) {
        Line line(m, b);
        roots[r] = (add(roots[r], line, x_min, x_max));
    }

    int64_t query(int r, int64_t x) {
        return query(roots[r], x, x_min, x_max);
    }

    int copy(int r) {
        roots.emplace_back(roots[r]);
        return roots.size() - 1;
    }

    int copy(int r1, int r2) {
        roots[r2] = roots[r1];
        return r2;
    }

    ~PersistantLiChao() {}
};

ll N, S[100005], V[100005], dp[100005];
vector<pll> T[100005];
PersistantLiChao plc(100005, 0, 1e9);

void dfs(int u, int p, ll pre) {
    if (u == 1) {
        dp[u] = 0;
        plc.add(u, 0, 0);
    }
    else {
        dp[u] = -plc.query(u, V[u]) + S[u] + pre * V[u];
        plc.add(u, pre, -dp[u]);
    }
    for (auto [v, w] : T[u])
        if (v != p) {
            plc.copy(u, v);
            dfs(v, u, pre + w);
        }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    plc.nodes.reserve(5000005);

    cin >> N;
    for (int i = 1; i < N; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        T[u].emplace_back(v, w);
        T[v].emplace_back(u, w);
    }
    for (int i = 2; i <= N; i++)
        cin >> S[i] >> V[i];
    
    dfs(1, 0, 0);

    for (int i = 2; i <= N; i++)
        cout << dp[i] << ' ';
    cout << '\n';
}

• Subtask 1: 40 / 100