这是100pts代码 但是疑惑点在于insert函数中if ( lca == sta[tp] ) return;这一句话为什么不用加上一句sta[++tp] = u;(也就是将当前处理的节点加进最右链中)就可以return了呢

题解中对此讲道 "首先，每插入一个点，如果栈顶元素是他所在链上的点，那么就可以不加这个点，这个画画图就知道了"

那加上这个点为啥只有20pts了呢/kk/kk/kk

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 5e5 + 5;
const int inf = 0x3f3f3f3f;
inline int read ()
{
    int x = 0 , f = 1;
    char ch = getchar();
    while ( !isdigit ( ch ) ) { if ( ch == '-' ) f = -1; ch = getchar (); }
    while ( isdigit ( ch ) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar (); }
    return x * f;
}

int n , m , a[N] , mn[N];//岛屿数量 敌方机器使用的次数 每一次的关键点编号
//最后一个是辅助数组 表示i到根路径上的最小边权

struct node 
{
    int head[N] , cnt;
    struct edge { int to , nxt , w; } e[N<<1];
    void add ( int u , int v , int w = 0 ) { e[++cnt] = { v , head[u] , w }; head[u] = cnt; }
}e1,e2;//e1是原来树的边 e2是现在虚树的边

int sz[N] , dep[N] , fa[N] , top[N] , son[N] , pos[N] , rev[N] , timer;
int tp , sta[N];

void dfs1 ( int u , int f )
{
    sz[u] = 1 , fa[u] = f , dep[u] = dep[f] + 1;
    for ( int i = e1.head[u] ; i ; i = e1.e[i].nxt )
    {
        int v = e1.e[i].to;
        if ( v == f ) continue;
        mn[v] = min ( mn[u] , e1.e[i].w );//这个v节点到根节点路径上的最小边权(也就是将v这个子树整个切掉所需要的最小代价)
        dfs1 ( v , u );
        sz[u] += sz[v];
        if ( sz[v] > sz[son[u]] ) son[u] = v;       
    }
}

void dfs2 ( int u , int tp )
{
    pos[u] = ++timer , rev[timer] = u , top[u] = tp;
    if ( son[u] ) dfs2 ( son[u] , tp );
    for ( int i = e1.head[u] ; i ; i = e1.e[i].nxt )
    {
        int v = e1.e[i].to;
        if ( v == fa[u] || v == son[u] ) continue;
        dfs2 ( v , v );
    }
}

int LCA ( int u , int v )
{
    while ( top[u] != top[v] )
    {
        if ( dep[top[u]] < dep[top[v]] ) swap ( u , v );//优先跳链顶深度大的节点
        u = fa[top[u]];
    }
    return dep[u] > dep[v] ? v : u;
}

void insert ( int u )
{
    if ( tp == 1 ) { if ( u != 1 ) sta[++tp] = u; return; }
    int lca = LCA ( u , sta[tp] );
    if ( lca == sta[tp] ) return;
    while ( tp > 1 && pos[sta[tp-1]] >= pos[lca] ) e2.add ( sta[tp-1] , sta[tp] ) , tp --;
    if ( lca != sta[tp] ) e2.add ( lca , sta[tp] ) , sta[tp] = lca;
    sta[++tp] = u;
}

void init()
{
    sta[tp=1] = 1;
    e2.cnt = 0;
}

int dfs ( int u )
{
    if ( !e2.head[u] ) return mn[u];
    int res = 0;
    for ( int i = e2.head[u] ; i ; i = e2.e[i].nxt )
    {
        int v = e2.e[i].to;
        res += dfs(v);
    }
    e2.head[u] = 0;
    return min ( res , mn[u] );
}

signed main ()
{
    n = read();
    memset ( mn , inf , sizeof ( mn ) );
    for ( int i = 1 , u , v , w ; i < n ; i ++ ) u = read() , v = read() , w = read() , e1.add ( u , v , w ) , e1.add ( v , u , w );
    dfs1 ( 1 , 0 ) , dfs2 ( 1 , 1 );
    m = read();
    for ( int i = 1 , k ; i <= m ; i ++ )
    {
        k = read();
        for ( int i = 1 ; i <= k ; i ++ ) a[i] = read();
        sort ( a + 1 , a + k + 1 , [](const int &a , const int &b) { return pos[a] < pos[b]; } );
        //按照dfs序排序
        init();
        for ( int i = 1 ; i <= k ; i ++ ) insert ( a[i] );
        for ( ; tp > 1 ; tp -- ) e2.add ( sta[tp-1] , sta[tp] );
        printf ( "%lld\n" , dfs(1) );
    }
    return 0;
}