(BEST) 定理，套用完成后，由于每一个路径都对应了 (deg_1) 这么多的不同起始方向的情况数，乘上去就可以了。

Code

#include<bits/stdc++.h>

using namespace std;

inline void read (int&a)
{
	a = 0; char k = getchar(); int f = 1;
	while (k > '9' || k < '0') { if (k == '0') f = -1; k = getchar(); }
	while (k >= '0' && k <= '9') { a = a * 10 + k - '0'; k = getchar (); }
	a *= f;
}

const int N = 200 + 5, mod = 1000003;
inline int plu (int u, int v) { return u + v >= mod ? u + v - mod : u + v; }
inline int sub (int u, int v) { return u - v < 0 ? u - v + mod : u - v; }
inline int mul (int u, int v) { return 1LL * u * v % mod; }
int a[N][N], n, outdeg[N], indeg[N], fac[200000 + 5];

inline int Guass ()
{
//	for (int i = 1; i <= n; ++i, puts (""))
//		for (int j = 1; j <= n; ++j)
//			printf ("%d ", a[i][j]);
	int ans = 1;
	for (int i = 1; i <= n - 1; ++i)
	{
		for (int j = i + 1; j <= n - 1; ++j)
		{
			while (a[j][i])
			{
				int l = a[i][i] / a[j][i];
				for (int k = i; k <= n - 1; ++k)
					a[i][k] = sub (a[i][k], mul (l, a[j][k]));
				swap (a[i], a[j]);
				ans = mod - ans;
			}
		}
		ans = mul (ans, a[i][i]);
	}
	return ans;
}

int main ()
{
	fac[0] = 1;
	for (int i = 1; i <= 200000; ++i)
		fac[i] = 1LL * fac[i - 1] * i % mod;
	while (scanf ("%d", &n) == 1 && n)
	{
		memset (a, 0, sizeof a);
		memset (outdeg, 0, sizeof outdeg);
		memset (indeg, 0, sizeof indeg);
		for (int i = 1; i <= n; ++i)
		{
			int s; read (s);
			for (int j = 1; j <= s; ++j)
			{
				int k; read (k);//i -> k
				a[i][k]--;
				a[i][i]++;
				indeg[k]++;
				outdeg[i]++;
			}
		}
		if (n == 1) { printf ("%d
", fac[ indeg[1] ]); continue; }//不用特判也可以过
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j)
				a[i][j] = sub (a[i][j], 0);
		bool flag = false;
		for (int i = 1; i <= n; ++i)
			if (indeg[i] != outdeg[i])
				{ printf ("0
"), flag = true; break; }
		if (flag) continue;
		int ans = 1;
		for (int i = 1; i <= n; ++i)
			ans = mul (ans, fac[ indeg[i] - 1 ]);
		ans = mul (ans, mul (Guass(), indeg[1]));
		printf ("%d
", ans);
	}
	return 0;
}

Conclusion

又开始错各种细节了。

比如矩阵忘了去掉某一行，而是直接 (n) 行计算。

注意某一行我写了不用特判也可以，因为高斯消元里面返回的因该是 (1) 。