不在场的证明(Checking an Alibi) 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>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int f, p, c;
    long long m;
    if (!(cin >> f >> p >> c >> m)) return 0;
    vector<vector<pair<int, long long>>> adj(f + 1);
    for (int i = 0; i < p; ++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(f + 1, INF);
    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]) {
                dist[v] = nd;
                pq.push({nd, v});
            }
        }
    }
    vector<int> ans;
    for (int i = 1; i <= c; ++i) {
        int pos;
        cin >> pos;
        if (dist[pos] <= m) ans.push_back(i);
    }
    cout << ans.size() << "\n";
    for (int x : ans) cout << x << "\n";
    return 0;
}