PS: 强连通，缩点。注意不要忘记考虑图是强连通的情况，WA了4次。省赛热身。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <stack>
using namespace std;
const int maxn = 110;
vector<int> G[maxn];
vector<int> G1[maxn];
vector<int> RG1[maxn];
int n;
int low[maxn], pre[maxn], sccno[maxn];
int dfs_clock, scc_cnt;
stack<int> S;
void dfs(int u) {
low[u] = pre[u] = ++dfs_clock;
S.push(u);
for(int i = 0; i < (int)G[u].size(); i++) {
int v = G[u][i];
if(!pre[v]) {
dfs(v);
low[u] = min(low[u], low[v]);
} else if(!sccno[v]) {
low[u] = min(low[u], low[v]);
}
}
if(pre[u]==low[u]) {
scc_cnt++;
for(;;) {
int x = S.top(); S.pop();
sccno[x] = scc_cnt;
if(x==u) break;
}
}
}
void find_scc() {
while(!S.empty()) S.pop();
memset(sccno, 0, sizeof(sccno));
memset(pre, 0, sizeof(pre));
memset(low, 0, sizeof(low));
scc_cnt = 0; dfs_clock = 0;
for(int i = 1; i <= n; i++) {
if(!pre[i]) dfs(i);
}
}
void build_DAG() {
int s, t;
for(int i = 1; i <= scc_cnt; i++) {
G1[i].clear();
RG1[i].clear();
}
for(int i = 1; i <= n; i++) {
for(int j = 0; j < (int)G[i].size(); j++) {
int v = G[i][j];
s = sccno[i]; t = sccno[v];
if(s!=t) {
G1[s].push_back(t);
RG1[t].push_back(s);
}
}
}
}
void work() {
int ans1 = 0, ans2 = 0;
for(int i = 1; i <= scc_cnt; i++) {
if(!RG1[i].size()) {
ans1++;
}
}
for(int i = 1; i <= scc_cnt; i++) {
if(!G1[i].size()) {
ans2++;
}
}
if(scc_cnt==1) { // WA.
printf("1\n0\n");
return;
}
printf("%d\n", ans1);
printf("%d\n", max(ans1, ans2));
}
int main()
{
int x;
while(scanf("%d", &n)!=EOF) {
for(int i = 1; i <= n; i++) G[i].clear();
for(int i = 1; i <= n; i++) {
for(;;) {
scanf("%d", &x);
if(x!=0) G[i].push_back(x);
else break;
}
}
find_scc();
build_DAG();
work();
}
return 0;
}