[BZOJ3611][HEOI2014]大工程（虚树+树形DP）

看到这样的问题，还是先构建出虚树。以下dep[u]为u在原树中的深度。
对于求最短距离和最长距离，实际上就类似于求最短链和最长链的问题，记一个最小（大）值和一个次小（大）值进行DP求出。
求边权和，记size[u]为u的子树内关键点的个数，f[u]为u的子树内所有关键点在原树中的深度之和，
g[u]=size[u](size[u]−1)2（也就是u的子树内关键点的对数）
h[u]=(size[u]−1)f[u]（可以理解为u的子树内所有关键点对的深度和）
以上四个值可以一遍DFS算出。
那么可以得出，LCA为u的关键点对数cnt[u]=g[u]−∑v∈son[u]g[v]。
LCA为u的所有关键点对的深度和sum[u]=h[u]−∑v∈son[u]h[v]。
由于点对(u,v)的距离为u的深度+v的深度−2∗LCA(u,v)的深度，因此以u为LCA的点对距离和为sum[u]−2∗cnt[u]∗dep[u]。
此题细节较多。
代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;inline int read() {    int res = 0; bool bo = 0; char c;    while (((c = getchar()) < '0' || c > '9') && c != '-');    if (c == '-') bo = 1; else res = c - 48;    while ((c = getchar()) >= '0' && c <= '9')        res = (res << 3) + (res << 1) + (c - 48);    return bo ? ~res + 1 : res;}typedef long long ll;const int N = 1e6 + 5, M = 2e6 + 5, LogN = 22, INF = 0x3f3f3f3f;int n, ecnt, nxt[M], adj[N], go[M], dep[N], fa[N][LogN], vir[N], vn, stk[N],par[N], vdn, dfn[N], times, aux[N], ecnt2, nxt2[M], adj2[N], go2[M],Max, Min, sze[N], maxd[N], mind[N]; ll sum, deps[N], dsum[N], dcnt[N];bool isvir[N];void add_edge(int u, int v) {    nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v;    nxt[++ecnt] = adj[v]; adj[v] = ecnt; go[ecnt] = u;}void add_edge2(int u, int v) {    nxt2[++ecnt2] = adj2[u]; adj2[u] = ecnt2; go2[ecnt2] = v;}void dfs(int u, int fu) {    int i; dep[u] = dep[fa[u][0] = fu] + 1; dfn[u] = ++times;    for (i = 0; i <= 19; i++)        fa[u][i + 1] = fa[fa[u][i]][i];    for (int e = adj[u], v; e; e = nxt[e])        if ((v = go[e]) != fu) dfs(v, u);}int lca(int u, int v) {    int i; if (dep[u] < dep[v]) swap(u, v);    for (i = 20; i >= 0; i--) {        if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];        if (u == v) return u;    }    for (i = 20; i >= 0; i--)        if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];    return fa[u][0];}bool comp(int u, int v) {    return dfn[u] < dfn[v];}void build() {    int i, top = 0; vdn = vn;    sort(vir + 1, vir + vn + 1, comp);    for (i = 1; i <= vdn; i++) {        int u = vir[i]; if (!top) {par[stk[++top] = u] = 0; continue;}        int w = lca(stk[top], u);        while (dep[stk[top]] > dep[w]) {            if (dep[stk[top - 1]] < dep[w]) par[stk[top]] = w;            top--;        }        if (w != stk[top]) vir[++vn] = w, par[w] = stk[top], stk[++top] = w;        par[stk[++top] = u] = w;    }    sort(vir + 1, vir + vn + 1, comp);    ecnt2 = 0; for (i = 1; i <= vn; i++) adj2[vir[i]] = 0;    for (i = 2; i <= vn; i++) add_edge2(par[vir[i]], vir[i]);}void dfs2(int u, int fu) {    sze[u] = isvir[u]; ll ds, dc; deps[u] = isvir[u] ? dep[u] : 0;    maxd[u] = -INF; mind[u] = INF; int nmax = -INF, nmin = INF;    for (int e = adj2[u], v; e; e = nxt2[e])        dfs2(v = go2[e], u), sze[u] += sze[v], deps[u] += deps[v];    ds = dsum[u] = 1ll * (sze[u] - 1) * deps[u];    dc = dcnt[u] = 1ll * sze[u] * (sze[u] - 1) >> 1;    for (int e = adj2[u], v; e; e = nxt2[e]) {        dc -= dcnt[v = go2[e]]; ds -= dsum[v];        if (maxd[v] > maxd[u]) nmax = maxd[u], maxd[u] = maxd[v];        else if (maxd[v] > nmax) nmax = maxd[v];        if (mind[v] < mind[u]) nmin = mind[u], mind[u] = mind[v];        else if (mind[v] < nmin) nmin = mind[v];    }    sum += ds - dc * dep[u] * 2;    if (maxd[u] != -INF && nmax != -INF)        Max = max(Max, maxd[u] + nmax - dep[u] * 2);    if (mind[u] != INF && nmin != INF)        Min = min(Min, mind[u] + nmin - dep[u] * 2);    if (isvir[u]) {        Max = max(Max, maxd[u] - dep[u]); Min = min(Min, mind[u] - dep[u]);        maxd[u] = max(maxd[u], dep[u]); mind[u] = min(mind[u], dep[u]);    }}int main() {    int i, x, y, q; n = read(); for (i = 1; i < n; i++) x = read(), y = read(),    add_edge(x, y); dfs(1, 0); q = read(); while (q--) {        vdn = vn = read(); for (i = 1; i <= vn; i++)            isvir[aux[i] = vir[i] = read()] = 1;        sum = 0; Max = -INF; Min = INF;        build(); dfs2(vir[1], 0); printf("%lld %d %d\n", sum, Min, Max);        for (i = 1; i <= vdn; i++) isvir[aux[i]] = 0;    }    return 0;}