香甜的黄油(Sweet Butter) 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;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, p, c;
    if (!(cin >> n >> p >> c)) return 0;
    vector<int> cows(p + 1, 0);
    for (int i = 0; i < n; ++i) {
        int x;
        cin >> x;
        cows[x]++;
    }
    vector<vector<pair<int, int>>> adj(p + 1);
    for (int i = 0; i < c; ++i) {
        int u, v, w;
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
    }
    const long long INF = (1LL << 58);
    long long ans = INF;
    for (int src = 1; src <= p; ++src) {
        vector<long long> dist(p + 1, INF);
        priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq;
        dist[src] = 0;
        pq.push({0, src});
        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]) {
                    dist[v] = nd;
                    pq.push({nd, v});
                }
            }
        }
        long long total = 0;
        bool ok = true;
        for (int i = 1; i <= p; ++i) {
            if (cows[i] > 0) {
                if (dist[i] >= INF) { ok = false; break; }
                total += dist[i] * cows[i];
            }
        }
        if (ok) ans = min(ans, total);
    }
    cout << ans << "\n";
    return 0;
}