奶牛派对(Cow Party S) 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;

vector<long long> dijkstra(int n, int s, const vector<vector<pair<int, int>>>& adj) {
    const long long INF = (1LL << 58);
    vector<long long> d(n + 1, INF);
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq;
    d[s] = 0;
    pq.push({0, s});
    while (!pq.empty()) {
        auto [du, u] = pq.top();
        pq.pop();
        if (du != d[u]) continue;
        for (auto& edge : adj[u]) {
            int v = edge.first;
            long long nd = du + edge.second;
            if (nd < d[v]) {
                d[v] = nd;
                pq.push({nd, v});
            }
        }
    }
    return d;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m, x;
    if (!(cin >> n >> m >> x)) return 0;
    vector<vector<pair<int, int>>> adj(n + 1), radj(n + 1);
    for (int i = 0; i < m; ++i) {
        int u, v, w;
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
        radj[v].push_back({u, w});
    }
    auto d1 = dijkstra(n, x, adj);
    auto d2 = dijkstra(n, x, radj);
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
        ans = max(ans, d1[i] + d2[i]);
    }
    cout << ans << "\n";
    return 0;
}