看标程的代码这么短，看我的....

难道是静态LCA模板太长了？ 

/*HDU 6074 - Phone Call [ LCA,并查集 ]  |  2017 Multi-University Training Contest 4题意：	给一棵树，定义集合S(u,v)为u到v路径上所有的点	给出 m 个 
     
       ，意思为集合里面的点互相距离为 w 	求 1 能到的所有点和该生成树的最小权值分析：	将所有线路按代价从小到大排序，对于每条线路(a,b,c,d)		分别把S(a,b)和S(c,d)合并到 LCA，最后再把两个 LCA 合并即可	再用f(i)表示i向上第一个与i不在同一个连通块的点, 就可用并查集压缩路径*/#include 
      
       using namespace std;#define LL long longconst int N = 100005;const int Q = 200005;vector
       
         G[N];namespace LCA{    struct Query {        int v, q;    }; vector 
        
          query[N];    int ans[Q], f[N], vis[N];    int sf(int x) {        return x == f[x] ? x : f[x] = sf(f[x]);    }    void init() {        memset(ans, -1, sizeof(ans));        for (int i = 0; i < N; i++) {            vis[i] = 0;            f[i] = i;            query[i].clear();        }    }    void adq(int u, int v, int id) {        query[u].push_back(Query{v, id});        query[v].push_back(Query{u, id});    }    void LCA(int u) {        f[u] = u;        vis[u] = 1;        for (auto& x : query[u]) {            if (vis[x.v] && ans[x.q] == -1)                ans[x.q] = sf(x.v);        }        for (auto& v : G[u]) {            if (vis[v]) continue;            LCA(v);            f[v] = u;        }    }}int pre[N];void dfs(int u, int fa) {    pre[u] = fa;    for (auto v : G[u]) {        if (v == fa) continue;        dfs(v, u);    }}int f[N], num[N];LL res[N];int sf(int x){    return x == f[x] ? x : f[x] = sf(f[x]);}int same[N];int Same(int x) {    return x == same[x] ? x : same[x] = Same(same[x]);}int w;void join(int a, int b){    a = sf(a), b = sf(b);    if (a == b) return;    f[b] = a;    num[a] += num[b];    res[a] += res[b] + w;}struct Node {    int a1, b1, a2, b2, w;    int c1, c2;}e[N];bool cmp(Node a, Node b) {    return a.w < b.w;}void solve(int a, int fa) {    while (Same(a) != Same(fa)) {        a = Same(a);        join(pre[a], a);        same[a] = pre[a];    }}int t, n, m;int main() {    scanf("%d", &t);    while (t--) {        for (int i = 0; i < N; i++) G[i].clear();        LCA::init();        scanf("%d%d", &n, &m);        for (int i = 1; i < n; i++) {            int u, v; scanf("%d%d", &u, &v);            G[u].push_back(v);            G[v].push_back(u);        }        int Q = 0;        for (int i = 1; i <= m; i++) {            scanf("%d%d%d%d%d", &e[i].a1, &e[i].b1, &e[i].a2, &e[i].b2, &e[i].w);            e[i].c1 = ++Q;            LCA::adq(e[i].a1, e[i].b1, Q);            e[i].c2 = ++Q;            LCA::adq(e[i].a2, e[i].b2, Q);        }        LCA::LCA(1);        for (int i = 1; i <= m; i++) {            e[i].c1 = LCA::ans[e[i].c1];            e[i].c2 = LCA::ans[e[i].c2];        }        dfs(1, 1);        sort(e+1, e+m+1, cmp);        for (int i = 1; i <= n; i++) {            f[i] = same[i] = i;            num[i] = 1;            res[i] = 0;        }        for (int i = 1; i <= m; i++) {            w = e[i].w;            solve(e[i].a1, e[i].c1);            solve(e[i].b1, e[i].c1);            solve(e[i].a2, e[i].c2);            solve(e[i].b2, e[i].c2);            join(e[i].c1, e[i].c2);        }        printf("%d %lld\n", num[sf(1)], res[sf(1)]);    }}