捷径(Shortcut 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 <numeric>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    long long T;
    if (!(cin >> n >> m >> T)) return 0;
    vector<long long> cows(n + 1);
    for (int i = 1; i <= n; ++i) cin >> cows[i];
    vector<vector<pair<int, long long>>> adj(n + 1);
    for (int i = 0; i < m; ++i) {
        int u, v;
        long long w;
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
    const long long INF = (1LL << 58);
    vector<long long> dist(n + 1, INF);
    vector<int> par(n + 1, 0);
    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;
        for (auto& edge : adj[u]) {
            int v = edge.first;
            long long nd = d + edge.second;
            if (nd < dist[v] || (nd == dist[v] && u < par[v])) {
                dist[v] = nd;
                par[v] = u;
                pq.push({nd, v});
            }
        }
    }
    vector<int> order(n);
    iota(order.begin(), order.end(), 1);
    sort(order.begin(), order.end(), [&](int a, int b) {
        if (dist[a] != dist[b]) return dist[a] > dist[b];
        return a > b;
    });
    for (int u : order) {
        if (u != 1 && par[u] > 0) {
            cows[par[u]] += cows[u];
        }
    }
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
        if (dist[i] > T) {
            ans = max(ans, (dist[i] - T) * cows[i]);
        }
    }
    cout << ans << "\n";
    return 0;
}