先枚举 $1\sim m$ 的每个节点，如果没被标记就 bfs 一次，bfs 的同时标记经过的每个节点，然后计算分数。每个节点排水排完之后都会清空

60pts，只通过了 $m = 1$ 的情况，剩下的都是 WA，代码：

//solve all tests
#include <iostream>
#include <vector>
#include <queue>

using namespace std;
#define int long long

const int N = 100010;

int n, m;
vector<int> v[N];
int d[N], dat1[N], dat2[N];
bool vis[N];
queue<int> q;

int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') { f = (ch == '-' ? -1 : f); ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
    return x * f;
}

int __gcd(int a, int b) {
    return b ? __gcd(b, a % b) : a;
}

int __lcm(int a, int b) {
    int gcd = __gcd(a, b);
    return a * b / gcd;
}

pair<int, int> make(int a, int b, int x, int y, int len) {
    int val1 = x, val2 = len * y;
    int gcd1 = __gcd(val1, val2);
    val1 /= gcd1, val2 /= gcd1;
    int mo = b * val2;
    int son = a * val2 + b * val1;
    int gcd = __gcd(mo, son);
    return make_pair(son / gcd, mo / gcd);
}

void bfs(int x) {
    q.push(x); vis[x] = true;
    while (q.size()) {
        int l = q.front(); q.pop();
        int len = v[l].size();
        if (!len) continue;
        for (int i = 0; i < len; i++) {
            int son = v[l][i]; vis[son] = true;
            // int lcm = __lcm(dat2[l], dat2[son]);
            // int x = lcm / dat2[l], y = lcm / dat2[son];
            // int val1 = dat1[son] * y + dat1[son] * x, val2 = lcm;
            // int val1 = dat1[son] + len * dat1[l], val2 = 
            // int gcd = __gcd(val1, val2);
            // dat1[son] = val1 / gcd, dat2[son] = val2 / gcd;
            pair<int, int> p = make(dat1[son], dat2[son], dat1[l], dat2[l], len);
            dat1[son] = p.first, dat2[son] = p.second;
            q.push(son);
        }
        dat1[l] = 0, dat2[l] = 1;
    }
}

signed main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> d[i];
        for (int j = 1; j <= d[i]; j++) {
            int x; cin >> x;
            v[i].push_back(x);
        }
        if (i <= m) dat1[i] = 1;
        dat2[i] = 1;
    }
    // for (int i = 1; i <= m; i++) dat1[i] = dat2[i] = 1;
    bfs(1);
    for (int i = 1; i <= m; i++)
        if (!vis[i]) bfs(i);
    for (int i = 1; i <= n; i++) {
        if (d[i]) continue;
        cout << dat1[i] << " " << dat2[i] << "\n";
    }
    return 0;
}