有N个岛和M座双向桥，编号为i的桥连接岛U[i]和V[i]，过桥耗时T[i]，桥连接两不同的岛屿，两个岛之间可能会有多座桥。
有Q组询问，每次询问给出K座桥，问从1号岛到N号岛的最少耗时，要求给出的K座桥分别至少经过1次。
2<=N<=400; N-1<=M<=2E5; 1<=U[i]<V[i]<=N; 1<=T[i]<=1E9; 1<=Q<=3000; 1<=K[i]<=5;
分析：跑floyd求出任意两岛之间的最短耗时，然后枚举所有可能的路径，即对K座桥全排列，并枚举桥的两端谁做入口谁做出口，将"1-桥-N"串起来得到耗时，取最小值。
#include <bits/stdc++.h>
using i64 = long long;const i64 inf = 1E18;
i64 d[405][405];void solve() {int N, M;std::cin >> N >> M;for (int i = 1; i <= N; i++) {for (int j = 1; j <= N; j++) {if (i == j) {d[i][j] = 0;} else {d[i][j] = inf;}}}std::vector<std::tuple<int,int,int>> edges(M+1);for (int i = 1; i <= M; i++) {int u, v;i64 t;std::cin >> u >> v >> t;edges[i] = {u, v, t};d[u][v] = std::min(d[u][v], t);d[v][u] = std::min(d[v][u], t);}for (int k = 1; k <= N; k++) {for (int i = 1; i <= N; i++) {for (int j = 1; j <= N; j++) {d[i][j] = std::min(d[i][j], d[i][k] + d[k][j]);}}}auto getb = [&](const std::vector<int> &v, int st, i64 &w) {std::vector<int> res;for (size_t i = 0; i < v.size(); i++) {int x, y, z;std::tie(x, y, z) = edges[v[i]];if (st & (1 << i)) {res.push_back(y);res.push_back(x);} else {res.push_back(x);res.push_back(y);}w += z;}return res;};int Q;std::cin >> Q;for (int i = 1; i <= Q; i++) {int K;std::cin >> K;std::vector<int> B(K);for (auto &x : B) {std::cin >> x;}std::sort(B.begin(), B.end());i64 ans = inf;do {for (int j = 0; j < (1 << K); j++) {i64 res = 0;auto b = getb(B, j, res);res += d[1][b.front()] + d[b.back()][N];for (size_t k = 2; k < b.size(); k += 2) {res += d[b[k]][b[k-1]];}ans = std::min(ans, res);}} while (std::next_permutation(B.begin(), B.end()));std::cout << ans << "\n";}
}int main() {std::cin.tie(0)->sync_with_stdio(0);int t = 1;while (t--) solve();return 0;
}
