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样例1对了，样例2对了一半，交上去一分都没有。

样例2：

输入

10 1
5 2 3 4 5 6
2 7 8
2 8 10
2 9 7
1 9
3 7 8 9
1 10
0
1 10
0

输出

4 15
11 15

我的输出

4 15
43 30

#include <bits/stdc++.h>
#define _for(i, a, b)  for (int i = (a); i <= (b); i ++ )
using namespace std;
const int N = 1e5 + 5;
typedef long long ll;
int n, m, vis[N];
vector<int> G[N];
struct node { ll son, mot; } water[N];
inline ll gcd(ll x, ll y) { return (y == 0 ? x : gcd(y, x % y)); }
inline ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
inline void add(ll & son, ll & mot, ll add_son, ll add_mot) {
	ll Gcd = gcd(add_son, add_mot);
    add_son /= Gcd, add_mot /= Gcd;
    ll new_mot = lcm(mot, add_mot), new_son = son * (new_mot / mot) + add_son * (new_mot / add_mot);
    ll GCd = gcd(new_son, new_mot);
    son = new_son / GCd, mot = new_mot / GCd;
}
void dfs(int x) { for (int y : G[x])  add(water[y].son, water[y].mot, water[x].son, water[x].mot * G[x].size()), dfs(y); }
inline void bfs() {
	queue<int> q;
	_for (i, 1, m)  q.push(i), vis[i] = 1;
	while (! q.empty()) {
		int x = q.front(); q.pop();
		for (int y : G[x])  if (! vis[y])  vis[y] = 1, add(water[y].son, water[y].mot, water[x].son, water[x].mot * G[x].size()), q.push(y);
	}
}
int main() {
    ios :: sync_with_stdio(false), cin.tie(0);
    cin >> n >> m;
    _for (i, 1, n)  water[i].son = 0, water[i].mot = 1;
    _for (i, 1, n) {
        int d, t;
        cin >> d;
        while (d -- )  cin >> t, G[i].push_back(t);
    }
    _for (i, 1, m)  water[i].son = 1;
    _for (i, 1, m)  dfs(i);
//    bfs();
    _for (i, m + 1, n)  if (! G[i].size())  cout << water[i].son << " " << water[i].mot << endl;
    return 0;
}