[SDOI2011]消耗战

题意：

给出n个点的一棵带有边权的树,以及q个询问.每次询问给出k个点,询问这使得这k个点与1点不连通所需切断的边的边权和最小是多少.

题解：

树型dp+虚树
dp[x]:切断x及其子树上询问点的最小代价
预处理出minv[pos]代表从11到pos路径上最小的边权
如果pos是询问点,dp(pos)=minv[pos]
否则，最小代价dp(pos)=min(minv[pos],∑dp(to))（其中to是pos的儿子）
如果pos为询问点，按理说不用dp[to]的值，但是仍然要对其儿子进行dfs，因为清空虚树需要对整个虚树进行遍历
如果对整个子树进行dp，复杂度过高，这时候就需要建虚树，（关于虚树见博文）
建虚图：

void insert(int x) {
    if(top == 1) {s[++top] = x; return ;}
    int lca = LCA(x, s[top]);
    if(lca == s[top]) return ;//以为s[top]也是关键点，那么s[top]子树里的点就没必要处理了 
    while(top > 1 && dfn[s[top - 1]] >= dfn[lca]) add_edge(s[top - 1], s[top]), top--;
    if(lca != s[top]) add_edge(lca, s[top]), s[top] = lca;//
    s[++top] = x;
}

为什么(lca == s[top]直接推出不把x加栈内
因为s[top]也是关键点，x也是关键点，x是s[top]的子树，那根据题意如果将s[top]与根节点断开，x节点自然也就断开了，也就是我们只需要考虑s[top]即可（x自动被考虑其中）

代码：

// luogu-judger-enable-o2
// luogu-judger-enable-o2
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
#define LL long long
char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 250001;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
char obuf[1 << 24], *O=obuf;
void print(LL x) {
    if(x > 9) print(x / 10);
    *O++= x % 10 + '0';
}
int N, M;
struct Edge {
    int u, v, w, nxt;
}E[MAXN << 1];
int head[MAXN], num = 1;
inline void AddEdge(int x, int y, int z) {
    E[num] = (Edge) {x, y, z, head[x]};
    head[x] = num++;
}
vector<int> v[MAXN];
void add_edge(int x, int y) {
    v[x].push_back(y);
}
int a[MAXN], dfn[MAXN], topf[MAXN], siz[MAXN], son[MAXN], s[MAXN], top, deep[MAXN], fa[MAXN], ID = 0;
LL mn[MAXN];
void dfs1(int x, int _fa) {
    siz[x] = 1; fa[x] = _fa;
    for(int i = head[x]; i != -1; i = E[i].nxt) {
        if(E[i].v == _fa) continue;
        deep[E[i].v] = deep[x] + 1;
        mn[E[i].v] = min(mn[x], (LL)E[i].w);
        dfs1(E[i].v, x);
        siz[x] += siz[E[i].v];
        if(siz[E[i].v] > siz[son[x]]) son[x] = E[i].v;
    }
}
void dfs2(int x, int topfa) {
    topf[x] = topfa;
    dfn[x] = ++ID;
    if(!son[x]) return ;
    dfs2(son[x], topfa);
    for(int i = head[x]; i != -1; i = E[i].nxt) 
        if(!topf[E[i].v]) 
            dfs2(E[i].v, E[i].v);
}
int LCA(int x, int y) {
    while(topf[x] != topf[y]) {
        if(deep[topf[x]] < deep[topf[y]]) swap(x, y);
        x = fa[topf[x]];
    }
    if(deep[x] < deep[y]) swap(x, y);
    return y;
}
void insert(int x) {
    if(top == 1) {s[++top] = x; return ;}
    int lca = LCA(x, s[top]);
    if(lca == s[top]) return ;//以为s[top]也是关键点，那么s[top]子树里的点就没必要处理了 
    while(top > 1 && dfn[s[top - 1]] >= dfn[lca]) add_edge(s[top - 1], s[top]), top--;
    if(lca != s[top]) add_edge(lca, s[top]), s[top] = lca;//
    s[++top] = x;
}
LL DP(int x) {
    if(v[x].size() == 0) return mn[x];
    LL sum = 0;
    for(int i = 0; i < v[x].size(); i++) 
        sum += DP(v[x][i]);
    v[x].clear();
    return min(sum, (LL)mn[x]);
}
int comp(const int &a, const int &b) {
    return dfn[a] < dfn[b];
}
int main() {
    memset(head, -1, sizeof(head));
    //memset(mn, 0xff, sizeof(mn));
    mn[1] = 1ll << 60;
    N = read();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read(), z = read();
        AddEdge(x, y, z); AddEdge(y, x, z);
    }
    deep[1] = 1;
    dfs1(1, 0);
    dfs2(1, 1);
    M = read();
    /*for(int i = 1; i <= N; i++)    
        for(int j = 1; j <= N; j++)
            printf("%d %d %d\n", i, j, LCA(i, j));*/
    //for(int i = 1; i <= N; i++) printf("%d ", mn[i]); puts("");
    while(M--) {
        int K = read();
        for(int i = 1; i <= K; i++) a[i] = read();
        sort(a + 1, a + K + 1, comp);
        s[top = 1] = 1;
        for(int i = 1; i <= K; i++) insert(a[i]);
        while(top > 0)  add_edge(s[top - 1], s[top]), top--;
        print(DP(1)), *O++ = '\n'; 
    }
    fwrite(obuf, O-obuf, 1 , stdout);    
    return 0;
}