最短路。

可以建立 $(0-9)$ $10$ 个超级源点，红绿灯编号有什么数字就建立有向边指向对应源点边权如题，再建立有向边从源点指向该红绿灯边权为 $0$。正常跑最短路，最后选出一家早餐店即可。

#include <bits/stdc++.h>

using i64 = long long;
using u32 = unsigned;
using u64 = unsigned long long;

void solve() {
    int n;
    std::cin >> n;

    std::vector<std::vector<std::pair<int, int>>> e(n + 12);
    for (int i = 0; i <= n; ++i) {
        int w;
        std::cin >> w;
        e[i].emplace_back(i + 1, w);
        e[i + 1].emplace_back(i, w);
    }

    for (int i = 0; i < n; ++i) {
        int u;
        std::cin >> u;

        while (u) {
            int t = u % 10;
            e[i + 1].emplace_back(n + 2 + t, t);
            e[n + 2 + t].emplace_back(i + 1, 0);
            u /= 10;
        }
    }

    std::vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) {
        std::cin >> a[i];
    }

    auto dijkstra = [&](int s) -> std::vector<i64> {
        std::vector<i64> dis(n + 12, 1E18);
        std::vector<int> vis(n + 12);
        std::priority_queue<std::pair<i64, int>, std::vector<std::pair<i64, int>>, std::greater<>> pq;
        dis[s] = 0;
        pq.emplace(0, s);
        while (!pq.empty()) {
            auto [_, u] = pq.top();
            pq.pop();

            if (vis[u]) {
                continue;
            }
            vis[u] = 1;

            for (auto [v, w] : e[u]) {
                if (dis[v] > _ + w) {
                    dis[v] = _ + w;
                    pq.emplace(dis[v], v);
                }
            }
        }
        return dis;
    };

    auto dis1 = dijkstra(0), dis2 = dijkstra(n + 1);
    i64 dis = 2E18, score = 0;
    for (int i = 1; i <= n; ++i) {
        i64 t = dis1[i] + dis2[i];
        if (t < dis) {
            dis = t;
            score = a[i];
        } else if (t == dis && a[i] > score) {
            score = a[i];
        }
    }

    std::cout << dis << " " << score << "\n";
}

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

    int t;
    std::cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}