Codeforces2014E Rendez-vous de Marian et Robin(分層圖最短路)

題意

有一個旅行網路，由 \(n\) 個點和 \(m\) 條邊組成。第 \(i\) 條邊連線點 \(u_i\)、\(v_i\)。

題解

容易想到

點選檢視程式碼

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

constexpr i64 inf = 1e18;

void solve() {
    int n, m, h;
    cin >> n >> m >> h;

    vector<vector<pair<int, int>>> adj(2 * n);
    for (int i = 0; i < h; i++) {
        int a;
        cin >> a;
        a--;
        
        adj[a].emplace_back(a + n, 0);
    }
    for (int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        u--; v--;

        adj[u].emplace_back(v, w);
        adj[v].emplace_back(u, w);
        adj[u + n].emplace_back(v + n, w / 2);
        adj[v + n].emplace_back(u + n, w / 2);
    }

    auto dijkstra = [&](int s) {
        priority_queue<pair<i64, int>, vector<pair<i64, int>>, greater<pair<i64, int>>> Q;
        vector<i64> dist(2 * n, inf);
        vector<bool> vis(2 * n, false);

        dist[s] = 0LL;
        Q.emplace(0LL, s);
        while (!Q.empty()) {
            int u = Q.top().second;
            Q.pop();

            if (vis[u]) {
                continue;
            }
            vis[u] = true;
            for (auto [v, w] : adj[u]) {
                if (dist[v] > dist[u] + w) {
                    dist[v] = dist[u] + w;
                    Q.emplace(dist[v], v);
                }
            }
        }
        
        vector<i64> ans(n);
        for (int i = 0; i < n; i++) {
            ans[i] = min(dist[i], dist[i + n]);
        }
        return ans;
    };

    vector<i64> D1 = dijkstra(0), D2 = dijkstra(n - 1);
    i64 ans = inf;
    for (int i = 0; i < n; i++) {
        ans = min(ans, max(D1[i], D2[i]));
    }

    if (ans == inf) {
        ans = -1;
    }
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    cin >> T;

    while (T--) {
        solve();
    }
    return 0;
}