题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6074

题意：给你一棵树，然后给你M个条件，每次给出a,b,c,d,cost,表示从a-->b,c-->d的路径中的点，可以互相到达，花费是cost，到达具有传递性 ，现在问你从1节点最多可以到达哪些节点，最小花费是多少。

解法：看着官方题解学的。


#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
typedef long long LL;
int T,n,m,parent[maxn],up[maxn],cnt[maxn];
struct FastIO
{
    static const int S = 1310720;
    int wpos;
    char wbuf[S];
    FastIO() : wpos(0) {}
    inline int xchar()
    {
        static char buf[S];
        static int len = 0, pos = 0;
        if (pos == len)
            pos = 0, len = fread(buf, 1, S, stdin);
        if (pos == len) return -1;
        return buf[pos ++];
    }
    inline LL Lint()
    {
        int c = xchar();
        LL x = 0;
        while (c <= 32) c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x;
    }
    inline int xint()
    {
        int s = 1, c = xchar(), x = 0;
        while (c <= 32) c = xchar();
        if (c == '-') s = -1, c = xchar();
        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
        return x * s;
    }
    inline void xstring(char *s)
    {
        int c = xchar();
        while (c <= 32) c = xchar();
        for (; c > 32; c = xchar()) * s++ = c;
        *s = 0;
    }
    inline void wchar(int x)
    {
        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
        wbuf[wpos ++] = x;
    }
    inline void wint(LL x)
    {
        if (x < 0) wchar('-'), x = -x;
        char s[24];
        int n = 0;
        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;
        while (n--) wchar(s[n]);
    }
    inline void wstring(const char *s)
    {
        while (*s) wchar(*s++);
    }
    ~FastIO()
    {
        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
    }
} io;
//parent[i]表示父亲节点是谁
//up[i]表示i点往上深度最大的一个可能不是和i在同一个连通块的祖先
//cnt[i]表示i所在连通块的点数
LL w[maxn]; //i所在连通块离的距离和
int dep[maxn],fa[maxn][20];
//dep[i]表示i点的深度,fa[i][j]代表i跳2^j的点
vector <int> G[maxn];
struct node{
    int a,b,c,d;
    LL cost;
    bool operator < (const node &rhs) const{
        return cost < rhs.cost;
    }
}q[maxn];//存下询问,按v值排序
void dfs1(int x, int d, int pre){
    dep[x] = d;
    fa[x][0] = pre;
    for(int i=1; i<20; i++){
        fa[x][i] = fa[fa[x][i-1]][i-1];
    }
    for(int i=0; i<G[x].size(); i++){
        int v=G[x][i];
        if(v == pre) continue;
        dfs1(v, d+1, x);
    }
}

int findfa(int x){
    if(x==parent[x]) return x;
    else return parent[x] = findfa(parent[x]);
}

int findup(int x){
    if(x == up[x]) return x;
    else return up[x] = findup(up[x]);
}

int LCA(int u, int v){
    if(dep[u]<dep[v]) swap(u,v);
    for(int i=19; i>=0; i--){
        if(dep[fa[u][i]]>=dep[v]){
            u=fa[u][i];
        }
    }
    if(u==v) return u;
    for(int i=19; i>=0; i--){
        if(fa[u][i]!=fa[v][i]){
            u=fa[u][i];
            v=fa[v][i];
        }
    }
    return fa[u][0];
}

void Union(int u, int v, LL cost)
{
    u = findfa(u), v = findfa(v);
    if(u == v) return;
    parent[u] = v;
    cnt[v] += cnt[u];
    w[v] += w[u] + cost;
}

void Merge(int u, int v, LL cost)
{
    while(1){
        u=findup(u);
        if(dep[u]<=dep[v]) return;
        Union(u, fa[u][0], cost);
        up[u] = fa[u][0];
    }
}

void solve(node s){
    int _lca = LCA(s.a, s.b);
    Merge(s.a, _lca, s.cost);
    Merge(s.b, _lca, s.cost);
    _lca = LCA(s.c, s.d);
    Merge(s.c, _lca, s.cost);
    Merge(s.d, _lca, s.cost);
    Union(s.a, s.c, s.cost);
}

int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    T = io.xint();
    while(T--)
    {
        n = io.xint(), m = io.xint();
        for(int i=0; i<=n; i++) parent[i]=up[i]=i,cnt[i]=1,w[i]=0,G[i].clear();
        for(int i=1; i<n; i++){
            int u, v;
            u = io.xint();
            v = io.xint();
            G[u].push_back(v);
            G[v].push_back(u);
        }
        for(int i=1; i<=m; i++){
            //scanf("%d %d %d %d %lld", &q[i].a, &q[i].b, &q[i].c, &q[i].d, &q[i].cost);
            q[i].a = io.xint();
            q[i].b = io.xint();
            q[i].c = io.xint();
            q[i].d = io.xint();
            q[i].cost = io.Lint();
        }
        sort(q+1, q+m+1);
        dfs1(1, 1, 0);
        for(int i=1; i<=m; i++)
        {
            solve(q[i]);
        }
        printf("%d %lld\n", cnt[findfa(1)], w[findfa(1)]);
    }
    return 0;
}