贸易(Trade) 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 <tuple>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m, k;
    if (!(cin >> n >> m >> k)) return 0;
    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<vector<long long>> dist(n + 1, vector<long long>(4, INF));
    using State = tuple<long long, int, int>;
    priority_queue<State, vector<State>, greater<State>> pq;
    dist[1][0] = 0;
    pq.push({0, 1, 0});
    while (!pq.empty()) {
        auto [d, u, s] = pq.top();
        pq.pop();
        if (d != dist[u][s]) continue;
        for (auto& edge : adj[u]) {
            int v = edge.first;
            long long w = edge.second;
            if (d + w < dist[v][s]) {
                dist[v][s] = d + w;
                pq.push({dist[v][s], v, s});
            }
            if (!(s & 1) && d < dist[v][s | 1]) {
                dist[v][s | 1] = d;
                pq.push({dist[v][s | 1], v, s | 1});
            }
            if (!(s & 2) && d + 2 * w < dist[v][s | 2]) {
                dist[v][s | 2] = d + 2 * w;
                pq.push({dist[v][s | 2], v, s | 2});
            }
            if (s == 0 && d + w < dist[v][3]) {
                dist[v][3] = d + w;
                pq.push({dist[v][3], v, 3});
            }
        }
    }
    for (int i = 0; i < k; ++i) {
        int t;
        cin >> t;
        cout << (t == 1 ? 0 : dist[t][3]) << "\n";
    }
    return 0;
}