1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
//Autor: Bartłomiej Czarkowski
#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 7;
int n, m, a;
int blok[N];
int stp[N];
vector <int> v[N];
vector <int> r[N];
vector <int> poj[N];
int odp[N];

int wylicz() {
	int ret = 0;
	for (int i = 1; i <= n; ++i) {
		if (blok[i]) {
			continue;
		}
		for (int j = 1; j <= n; ++j) {
			stp[j] = v[j].size();
		}
		stp[i] = 0;
		queue <int> que;
		que.push(i);
		--ret;
		while (!que.empty()) {
			int x = que.front();
			que.pop();
			++ret;
			for (int j : r[x]) {
				if (blok[j]) {
					continue;
				}
				--stp[j];
				if (!stp[j]) {
					que.push(j);
				}
			}
		}
	}
	return ret;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &m);
		for (int j = 1; j <= m; ++j) {
			scanf("%d", &a);
			v[i].push_back(a);
			r[a].push_back(i);
		}
		if (v[i].size() == 1) {
			poj[v[i][0]].push_back(i);
		}
	}
	for (int i = n; i; --i) {
		odp[i] = wylicz();
		blok[i] = 1;
	}
	for (int i = 1; i <= n; ++i) {
		printf("%d ", odp[i]);
	}
	printf("\n");
	return 0;
}