旅游巴士(Tourist Bus) 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 a;
        cin >> u >> v >> a;
        adj[u].push_back({v, a});
    }
    const long long INF = (1LL << 58);
    vector<vector<long long>> dist(n + 1, vector<long long>(k, 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, r] = pq.top();
        pq.pop();
        if (d != dist[u][r]) continue;
        int nr = (r + 1) % k;
        for (auto& edge : adj[u]) {
            int v = edge.first;
            long long a = edge.second;
            long long nd = d;
            if (nd < a) {
                long long wait_k = (a - nd + k - 1) / k;
                nd += wait_k * k;
            }
            nd += 1;
            if (nd < dist[v][nr]) {
                dist[v][nr] = nd;
                pq.push({nd, v, nr});
            }
        }
    }
    if (dist[n][0] >= INF) cout << -1 << "\n";
    else cout << dist[n][0] << "\n";
    return 0;
}