树上买票旅行(Putovanje) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

struct Edge {
    int u, v;
    long long c1, c2;
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<Edge> edges(n - 1);
    vector<vector<int>> adj(n + 1);
    for (int i = 0; i < n - 1; ++i) {
        cin >> edges[i].u >> edges[i].v >> edges[i].c1 >> edges[i].c2;
        adj[edges[i].u].push_back(edges[i].v);
        adj[edges[i].v].push_back(edges[i].u);
    }
    const int LOG = 19;
    vector<vector<int>> up(LOG, vector<int>(n + 1, 0));
    vector<int> dep(n + 1, 0), order;
    order.reserve(n);
    queue<int> q;
    dep[1] = 1;
    q.push(1);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        order.push_back(u);
        for (int v : adj[u]) {
            if (!dep[v]) {
                dep[v] = dep[u] + 1;
                up[0][v] = u;
                q.push(v);
            }
        }
    }
    for (int k = 1; k < LOG; ++k) {
        for (int i = 1; i <= n; ++i) {
            up[k][i] = up[k - 1][up[k - 1][i]];
        }
    }
    auto lca = [&](int u, int v) {
        if (dep[u] < dep[v]) swap(u, v);
        int d = dep[u] - dep[v];
        for (int k = 0; k < LOG; ++k) {
            if ((d >> k) & 1) u = up[k][u];
        }
        if (u == v) return u;
        for (int k = LOG - 1; k >= 0; --k) {
            if (up[k][u] != up[k][v]) {
                u = up[k][u];
                v = up[k][v];
            }
        }
        return up[0][u];
    };
    vector<long long> diff(n + 1, 0);
    for (int i = 1; i < n; ++i) {
        int l = lca(i, i + 1);
        diff[i]++;
        diff[i + 1]++;
        diff[l] -= 2;
    }
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        if (up[0][u]) diff[up[0][u]] += diff[u];
    }
    long long ans = 0;
    for (int i = 0; i < n - 1; ++i) {
        int child = (dep[edges[i].u] > dep[edges[i].v] ? edges[i].u : edges[i].v);
        ans += min(edges[i].c1 * diff[child], edges[i].c2);
    }
    cout << ans << "\n";
    return 0;
}