思路：首先这道题目是一个一看就是要用分层图来做，层与层之间只有隧道将不同的层联系在一起，结合题意每次穿越隧道的转线的时候，还要等$w_i$分钟等下一班地铁的到来，这里我们选择每一站选择设置两个状态，一个是上车状态，一个是等车状态。然后对每个询问直接跑spfa。

#include <iostream>
#include <cstring>

using namespace std;

const int N = 2e5 + 1e4, M = N * 2;

int h[N], e[M], ne[M], w[M], idx, q[N], value[N], s_x, s_y, e_x, e_y, dist[N];
bool vis[N];

void add (int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void spfa ()
{
    memset (dist, 0x3f, sizeof dist);
    int hh = 0, tt = 0;
    q[tt ++] = s_x * 2000 + (s_y - 1) * 2;
    dist[s_x * 2000 + (s_y - 1) * 2] = 0;
    while (hh != tt)
    {
        int t = q[hh ++];
        if (hh == N) hh = 0;
        vis[t] = false;
        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j]  = dist[t] + w[i];
                if (!vis[j]) vis[j] = true, q[tt ++] = j;
                if (tt == N) tt = 0;
            }
        }
    }
}



int main()
{
    int T; scanf ("%d", &T);
    for (int Case = 1; Case <= T; Case ++)
    {
        memset (h, -1, sizeof h);
        idx = 0;
        int n; scanf ("%d", &n); // 线路数量。
        for (int i = 1; i <= n; i ++)
        {
            int a; scanf ("%d%d", &a, value + i);
            add (i * 2000, i * 2000 + 1, value[i]);
            for (int j = 1, k = 3; j <= a - 1; j ++, k += 2) // 加一是上车状态。
            {

                int times; scanf ("%d", &times);
                add (i * 2000 + k - 1, i * 2000 + k, value[i]);
                add (i * 2000 + k - 2, i * 2000 + k, times);
                add (i * 2000 + k, i * 2000 + k - 2, times);
            }

        }
        int m, cnt = 1; scanf ("%d", &m);
        for (int i = 1; i <= m; i ++, cnt ++)
        {

            int a, b, c, d, e; scanf ("%d%d%d%d%d", &a, &b, &c, &d, &e);
            add (a * 2000 + (b - 1) * 2 + 1, c * 2000 + (d - 1) * 2, e);
            add (a * 2000 + (b - 1) * 2, c * 2000 + 2 * (d - 1), e);
            add (c * 2000 + (d - 1) * 2, a * 2000 + (b - 1) * 2, e);
            add (c * 2000 + (d - 1) * 2 + 1, a * 2000 + (b - 1) * 2, e);

        }
        int Q; scanf ("%d", &Q);
        printf ("Case #%d:\n", Case);
        while (Q --)
        {
            scanf ("%d%d%d%d", &s_x, &s_y, &e_x, &e_y);
            spfa();
            int t = min(dist[e_x * 2000 + (e_y - 1) * 2], dist[e_x * 2000 + (e_y - 1) * 2 + 1]);
            if (t != 0x3f3f3f3f)printf ("%d\n", t);
            else puts ("-1");
        }
    }
    return 0;
}