1 #include <bits/stdc++.h> 2 3 using namespace std; 4 #define _for(i,a,b) for(int i = (a);i < b;i ++) 5 #define _rep(i,a,b) for(int i = (a);i > b;i --) 6 #define INF 0x3f3f3f3f 7 #define pb push_back 8 #define maxn 500 9 typedef long long ll; 10 inline ll read() 11 { 12 ll ans = 0; 13 char ch = getchar(), last = ' '; 14 while(!isdigit(ch)) last = ch, ch = getchar(); 15 while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar(); 16 if(last == '-') ans = -ans; 17 return ans; 18 } 19 inline void write(ll x) 20 { 21 if(x < 0) x = -x, putchar('-'); 22 if(x >= 10) write(x / 10); 23 putchar(x % 10 + '0'); 24 } 25 vector<int> G[maxn]; 26 vector<int> rG[maxn]; 27 vector<int> vs; 28 vector<int> ans[maxn]; 29 vector<int> G2[maxn]; 30 map<int,int> m; 31 bool vis[maxn]; 32 bool used[maxn]; 33 int V,E; 34 int k = 0; 35 void add_edge(int from,int to) 36 { 37 G[from].pb(to); 38 rG[to].pb(from); 39 } 40 void dfs(int v) 41 { 42 used[v] = true; 43 _for(i,0,G[v].size()) 44 if(!used[G[v][i]]) 45 dfs(G[v][i]); 46 vs.pb(v); 47 } 48 void rdfs(int v,int k) 49 { 50 used[v] = true; 51 m[v] = k; 52 ans[k].pb(v); 53 _for(i,0,rG[v].size()) 54 if(!used[rG[v][i]]) 55 rdfs(rG[v][i],k); 56 } 57 //第k个集合里有哪些节点 ans[k] 58 //rnt为最大集合的集合大小 59 void Kosaraju() 60 { 61 memset(used,0,sizeof(used)); 62 _for(v,1,V+1) 63 if(!used[v]) 64 dfs(v); 65 66 memset(used,0,sizeof(used)); 67 68 _rep(i,vs.size()-1,-1) 69 if(!used[vs[i]]) 70 { 71 rdfs(vs[i],k); 72 k ++; 73 } 74 } 75 int in = 0,out = 0; 76 int ans1 = 0,ans2 = 0; 77 void solve() 78 { 79 int a[maxn]; 80 memset(a,0,sizeof(a)); 81 _for(i,0,k) 82 { 83 if(!G2[i].size()) 84 out ++; 85 _for(j,0,G2[i].size()) 86 { 87 a[G2[i][j]] ++; 88 } 89 } 90 _for(i,0,k) 91 if(!a[i]) 92 in ++; 93 ans1 = in; 94 ans2 = max(in,out); 95 if(k==1) 96 ans2 = 0; 97 } 98 int main() 99 { 100 V = read(); 101 _for(i,1,V+1) 102 { 103 int t = read(); 104 while(t) 105 { 106 add_edge(i,t); 107 t = read(); 108 } 109 } 110 Kosaraju(); 111 112 _for(i,1,V+1) 113 { 114 if(!vis[i]) 115 { 116 int nwbl = m[i]; 117 int sz = ans[nwbl].size(); 118 _for(j,0,sz) 119 { 120 vis[ans[nwbl][j]] = 1; 121 _for(k,0,G[ans[nwbl][j]].size()) 122 if(nwbl!=m[G[ans[nwbl][j]][k]]) 123 G2[nwbl].pb(m[G[ans[nwbl][j]][k]]); 124 } 125 } 126 } 127 solve(); 128 printf("%d %d ",ans1,ans2); 129 return 0; 130 }