题目链接：校园网Network of Schools
这个题得翻译下题目意思才知道在干嘛，题目一开始表明了这个是一个有向图，因为边是单向的。其次关于第一个问题：
基于一个事实，如果有 \(x \rightarrow y \rightarrow z\)，那么只需要 \(x\) 接受协议，它所在的 \(scc\) 强连通分量上的点一定都能不需要接受协议了，那么其实做缩点以后，算入度为 \(0\) 的强联通分量数量就行了。
第二个问题，问还需要加多少条边完成互通，显然加到整体变成一个强连通分量就可以互通了。那么最少加几条边呢？如果原来本身就是强连通分量，那么显然不需要加边。否则考虑缩点以后：
如图所示，我们需要干一件事：
添加边连向出度为 \(0\) 的点与入度为 \(0\) 的点，使得它俩连通，至于怎么连无所谓。假设入度为 \(0\) 的点为 \(x\)，出度为 \(0\) 的点为 \(y\) 个。我们需要把所有的 \(x\) 和所有的 \(y\) 连接，连接方式随意，显然至少需要连 \(\max(x,y)\) 条边，当然如果本身是 \(scc\) 显然不需要连边，特判就行。至于求 \(scc\) 直接跑 \(tarjan\) 就行了，然后缩点，连边并不需要真的连，直接记录出入度变化就行了。
参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#define isPbdsFile
#ifdef isPbdsFile
#include <bits/extc++.h>
#else
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
#endif
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
template <typename T>
int disc(T* a, int n)
{
return unique(a + 1, a + n + 1) - (a + 1);
}
template <typename T>
T lowBit(T x)
{
return x & -x;
}
template <typename T>
T Rand(T l, T r)
{
static mt19937 Rand(time(nullptr));
uniform_int_distribution<T> dis(l, r);
return dis(Rand);
}
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
return (a % b + b) % b;
}
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
a %= c;
T1 ans = 1;
for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
return modt(ans, c);
}
template <typename T>
void read(T& x)
{
x = 0;
T sign = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-')sign = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
x *= sign;
}
template <typename T, typename... U>
void read(T& x, U&... y)
{
read(x);
read(y...);
}
template <typename T>
void write(T x)
{
if (typeid(x) == typeid(char))return;
if (x < 0)x = -x, putchar('-');
if (x > 9)write(x / 10);
putchar(x % 10 ^ 48);
}
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
write(x), putchar(c);
write(c, y...);
}
template <typename T11, typename T22, typename T33>
struct T3
{
T11 one;
T22 tow;
T33 three;
bool operator<(const T3 other) const
{
if (one == other.one)
{
if (tow == other.tow)return three < other.three;
return tow < other.tow;
}
return one < other.one;
}
T3() { one = tow = three = 0; }
T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
{
}
};
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
if (x < y)x = y;
}
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
if (x > y)x = y;
}
constexpr int N = 110;
vector<int> child[N];
int scc[N], sccCnt;
int dfn[N], low[N];
int n, cnt;
stack<int> st;
inline void tarjan(const int curr)
{
st.push(curr);
dfn[curr] = low[curr] = ++cnt;
for (const int nxt : child[curr])
{
if (!dfn[nxt])tarjan(nxt), uMin(low[curr], low[nxt]);
else if (!scc[nxt])uMin(low[curr], dfn[nxt]);
}
if (dfn[curr] == low[curr])
{
++sccCnt;
while (true)
{
const int nxt = st.top();
st.pop();
scc[nxt] = sccCnt;
if (nxt == curr)break;
}
}
}
int in[N], out[N];
inline void solve()
{
cin >> n;
forn(i, 1, n)
{
int son;
cin >> son;
while (son)child[i].push_back(son), cin >> son;
}
forn(i, 1, n)if (!dfn[i])tarjan(i);
forn(i, 1, n)
{
for (const int j : child[i])
{
if (scc[i] != scc[j])
{
//scc[i]->scc[j]
in[scc[j]]++;
out[scc[i]]++;
}
}
}
int ansA = 0, ansB = 0;
forn(i, 1, sccCnt)ansA += in[i] == 0, ansB += out[i] == 0;
uMax(ansB, ansA);
if (sccCnt == 1)ansB = 0;
cout << ansA << endl << ansB;
}
signed int main()
{
// MyFile
Spider
//------------------------------------------------------
// clock_t start = clock();
int test = 1;
// read(test);
// cin >> test;
forn(i, 1, test)solve();
// while (cin >> n, n)solve();
// while (cin >> test)solve();
// clock_t end = clock();
// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}