思路：分层图，第一层图的边权为均为w，第二层图的边权均为w / 2，第一层图的点i与第二层图的点i + n对应，由于骑上马之后一定比骑上马再下马更优，因此两层图之间用单向边连接，即，如果在点i上有马，那么就在i与i + n之间连一条边权为0的单向边。最后从点1和点n分别跑dijkstra即可。最后一个点的最短时间就是min(dis[i], dis[i + n])。
用数组模拟建边会超时 注意开longlong

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

void solve() {
    int n, m, h;
    cin >> n >> m >> h;
    vector<vector<pair<int, int>>> adj(2 * n + 1); 
    for(int i = 1; i <= h; i ++ ) {
        int x;
        cin >> x;
        adj[x].push_back({x + n, 0});
    }

    while(m -- ) {
        int u, v, w; 
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
        adj[v].push_back({u, w});
        adj[u + n].push_back({v + n, w / 2});
        adj[v + n].push_back({u + n, w / 2});
    }

    vector<bool> st(2 * n + 1);
    auto dijkstra = [&](int start, vector<ll> &dis) {
        dis[start] = 0;
        priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<>> q;
        q.push({0ll, start});
        while(!q.empty()) {
            auto t = q.top();
            q.pop();
            ll dist = t.first, ver = t.second;
            if(st[ver]) continue;
            st[ver] = true;

            for(auto t: adj[ver]) {
                if(dis[t.first] > dist + t.second) {
                    dis[t.first] = dist + t.second;
                    q.push({dis[t.first], t.first});
                }
            }
        }
    };

    vector<ll> dis(2 * n + 1, 1e18), redis(2 * n + 1, 1e18);
    dijkstra(1, dis);
    for(int i = 1; i <= 2 * n; i ++ ) st[i] = false;
    dijkstra(n, redis);
    ll ans = 1e18;
    for(int i = 1; i <= n; i ++ ) {
        ans = min(ans, max(min(dis[i], dis[i + n]), min(redis[i], redis[i + n])));
    }
    if(ans == 1e18) cout << "-1\n";
    else cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while(t -- ) solve();
    return 0;
}