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答案
#include <iostream>
#include <vector>
#include <cstring>
#include <cstdio>
using namespace std;
#define MAXV 1010
bool vis[MAXV];
pair<long long, int> cost_a2u[MAXV];
pair<long long, int> cost[MAXV][MAXV];
int main() {
freopen("cowroute.in", "r", stdin);
freopen("cowroute.out", "w", stdout);
int A, B, N;
cin >> A >> B >> N;
/* Initialize cost matrix to infinity. */
memset(cost, 0x3F, sizeof(cost));
for (int i = 0; i < MAXV; i++) {
cost[i][i] = make_pair(0, 0);
}
for (int i = 0; i < N; i++) {
long long route_cost;
int route_len;
cin >> route_cost >> route_len;
vector<int> route(route_len);
for (int j = 0; j < route_len; j++) {
cin >> route[j];
/* Update the cost from everything before this city to this city. */
for (int k = 0; k < j; k++) {
cost[route[k]][route[j]] = min(cost[route[k]][route[j]],
make_pair(route_cost, j - k));
}
}
}
/* Perform Dijkstra without a heap in O(V^2) time. */
memset(vis, 0, sizeof(vis));
memset(cost_a2u, 0x3F, sizeof(cost_a2u));
cost_a2u[A] = make_pair(0, 0);
for (int i = 0; i < MAXV; i++) {
/* Find the closest unvisited vertex. */
int u = -1;
for (int j = 0; j < MAXV; j++) {
if (vis[j]) {
continue;
} else if (u == -1 || cost_a2u[j] < cost_a2u[u]) {
u = j;
}
}
/* Relax vertex u. */
vis[u] = true;
for (int j = 0; j < MAXV; j++) {
pair<long long, int> rlx = cost_a2u[u];
rlx.first += cost[u][j].first;
rlx.second += cost[u][j].second;
cost_a2u[j] = min(cost_a2u[j], rlx);
}
}
/* Output the cheapest cost and length if possible. */
if (cost_a2u[B].second <= MAXV) {
cout << cost_a2u[B].first << ' ' << cost_a2u[B].second << endl;
} else {
cout << "-1 -1\n";
}
return 0;
}

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