牛奶泵(Milk Pumping G) 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 to, cost, flow;
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<vector<Edge>> adj(n + 1);
    vector<int> flows;
    for (int i = 0; i < m; ++i) {
        int u, v, c, f;
        cin >> u >> v >> c >> f;
        adj[u].push_back({v, c, f});
        adj[v].push_back({u, c, f});
        flows.push_back(f);
    }
    sort(flows.begin(), flows.end());
    flows.erase(unique(flows.begin(), flows.end()), flows.end());
    const long long INF = (1LL << 58);
    long long best = 0;
    for (int f_lim : flows) {
        vector<long long> dist(n + 1, INF);
        priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq;
        dist[1] = 0;
        pq.push({0, 1});
        while (!pq.empty()) {
            auto [d, u] = pq.top();
            pq.pop();
            if (d != dist[u]) continue;
            if (u == n) break;
            for (const auto& e : adj[u]) {
                if (e.flow < f_lim) continue;
                if (d + e.cost < dist[e.to]) {
                    dist[e.to] = d + e.cost;
                    pq.push({dist[e.to], e.to});
                }
            }
        }
        if (dist[n] < INF) {
            best = max(best, (1000000LL * f_lim) / dist[n]);
        }
    }
    cout << best << "\n";
    return 0;
}