第一行一个整数n，代表岛屿数量。

接下来n-1行，每行三个整数u,v,w，代表u号岛屿和v号岛屿由一条代价为c的桥梁直接相连，保证1<=u,v<=n且1<=c<=100000。

第n+1行，一个整数m，代表敌方机器能使用的次数。

接下来m行，每行一个整数k_i，代表第i次后，有k_i个岛屿资源丰富，接下来k个整数h₁,h₂,…h_k，表示资源丰富岛屿的编号。

输出有m行，分别代表每次任务的最小代价。

对于100%的数据，2<=n<=250000,m>=1,sigma(ki)<=500000,1<=ki<=n-1

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u]; k != -1; k = ed[k].nxt)
#define PP(a,b) printf("link %d -> %d\n",a,b)
using namespace std;
const int maxn = 250005,maxm = 510005;
const LL INF = 100000000000000000ll;
inline int RD(){int out = 0,flag = 1; char c = getchar();while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();}return out * flag;
}
int n,m,K,h[maxn],ne = 0,fa[maxn][20],dfn[maxn],dep[maxn],cnt = 0;
int s[maxn],t[maxn];
LL mn[maxn],f[maxn];
struct EDGE{int to,nxt; LL w;}ed[maxm];
inline void build(int u,int v,int w){ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;
}
inline void add(int u,int v){if (u != v) {ed[ne] = (EDGE){v,h[u],0}; h[u] = ne++;}}
void dfs(int u,int ff,int d){fa[u][0] = ff; dfn[u] = ++cnt; dep[u] = ++d;Redge(u) if (ed[k].to != ff){mn[ed[k].to] = min(mn[u],ed[k].w);dfs(ed[k].to,u,d);}
}
void init(){REP(j,19) REP(i,n) fa[i][j] = fa[fa[i][j - 1]][j - 1];}
int lca(int u,int v){if (dep[u] < dep[v]) swap(u,v);int d = dep[u] - dep[v];for (int i = 0; (1 << i) <= d; i++)if (d & (1 << i)) u = fa[u][i];for (int i = 19; i >= 0; i--)if (fa[u][i] != fa[v][i]) u = fa[u][i],v = fa[v][i];if (u == v) return u;return fa[u][0];
}
inline bool cmp(const int& a,const int& b){return dfn[a] < dfn[b];}
void rebuild(){int top,tot = 0; s[top = 1] = 1; ne = 0;sort(t + 1,t + 1 + K,cmp); t[++tot] = t[1];for (int i = 2 ; i <= K; i++) if (lca(t[i],t[tot]) != t[tot]) t[++tot] = t[i];for (int i = 1; i <= tot; i++){int u = t[i],v = lca(u,s[top]);while (true){if (dep[v] >= dep[s[top - 1]]){add(v,s[top--]); if (v != s[top]) s[++top] = v;break;}add(s[top - 1],s[top]); top--;}if (s[top] != u) s[++top] = u;}while (--top) add(s[top],s[top + 1]);
}
void dp(int u){f[u] = mn[u]; LL tmp = 0;Redge(u) dp(ed[k].to),tmp += f[ed[k].to];h[u] = -1;if (tmp == 0) f[u] = mn[u];else if (tmp <= f[u]) f[u] = tmp;
}
void solve(){K = RD(); REP(i,K) t[i] = RD();rebuild(); dp(1);printf("%lld\n",f[1]);
}
int main(){memset(h,-1,sizeof(h));n = RD(); int a,b,w;for (int i = 1; i < n; i++) a = RD(),b = RD(),w = RD(),build(a,b,w);mn[1] = INF; dfs(1,0,0); init();REP(i,n) h[i] = -1;m = RD(); while (m--) solve();return 0;
}