/*
Tarjan算法
复杂度O(N+M)
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;const int MAXN=20010;//点数
const int MAXM=50010;//边数
struct Edge{int to,next;
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~scc
int Index,top;
int scc;//强连通分量的个数
bool Instack[MAXN];
int num[MAXN];//各个强连通分量包含点的个数，数组编号1~scc
//num数组不一定需要，结合实际情况void addedge(int u,int v){edge[tot].to=v;edge[tot].next=head[u];head[u]=tot++;
}
void Tarjan(int u){int v;Low[u]=DFN[u]=++Index;Stack[top++]=u;Instack[u]=true;for(int i=head[u];i!=-1;i=edge[i].next){v=edge[i].to;if(!DFN[v]){Tarjan(v);if(Low[u]>Low[v])Low[u]=Low[v];}else if(Instack[v]&&Low[u]>DFN[v])Low[u]=DFN[v];}if(Low[u]==DFN[u]){scc++;do{v=Stack[--top];Instack[v]=false;Belong[v]=scc;num[scc]++;}while(v!=u);}
}
void solve(int N){memset(DFN,0,sizeof(DFN));memset(Instack,false,sizeof(Instack));memset(num,0,sizeof(num));Index=scc=top=0;for(int i=1;i<=N;i++)if(!DFN[i])Tarjan(i);
}
void init(){tot=0;memset(head,-1,sizeof(head));
}int main(){return 0;
}

/*
Kosaraju算法，复杂度O(N+M)
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;const int MAXN=20010;
const int MAXM=50010;
struct Edge{int to,next;
}edge1[MAXM],edge2[MAXM];
//edge1是原图，edge2是逆图GT
int head1[MAXN],head2[MAXN];
bool mark1[MAXN],mark2[MAXN];
int tot1,tot2;
int cnt1,cnt2;
int st[MAXN];//对原图进行dfs，点的结束时间从小到大排序
int Belong[MAXN];//每个点属于哪个连通分量（0~cnt2-1）
int num;//中间变量，用来数某个连通分量中点的个数
int setNum[MAXN];//强连通分量中点的个数，编号0~cnt2-1
void addedge(int u,int v){edge1[tot1].to=v;edge1[tot1].next=head1[u];head1[u]=tot1++;edge2[tot2].to=u;edge2[tot2].next=head2[v];head2[v]=tot2++;
}
void DFS1(int u){mark1[u]=true;for(int i=head1[u];i!=-1;i=edge1[i].next)if(!mark1[edge1[i].to])DFS1(edge1[i].to);st[cnt1++]=u;
}
void DFS2(int u){mark2[u]=true;num++;Belong[u]=cnt2;for(int i=head2[u];i!=-1;i=edge2[i].next)if(!mark2[edge2[i].to])DFS2(edge2[i].to);
}
//点的编号从1开始
void solve(int n){memset(mark1,false,sizeof(mark1));memset(mark2,false,sizeof(mark2));cnt1=cnt2=0;for(int i=1;i<=n;i++)if(!mark1[i])DFS1(i);for(int i=cnt1-1;i>=0;i--)if(!mark2[st[i]]){num=0;DFS2(st[i]);setNum[cnt2++]=num;}
}int main(){return 0;
}

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
/*
*  求 无向图的割点和桥
*  可以找出割点和桥，求删掉每个点后增加的连通块。
*  需要注意重边的处理，可以先用矩阵存，再转邻接表，或者进行判重
*/
const int MAXN = 10010;
const int MAXM = 100010;
struct Edge
{int to,next;bool cut;//是否为桥的标记
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN];
int Index,top;
bool Instack[MAXN];
bool cut[MAXN];
int add_block[MAXN];//删除一个点后增加的连通块
int bridge;void addedge(int u,int v)
{edge[tot].to = v;edge[tot].next = head[u];edge[tot].cut = false;head[u] = tot++;
}void Tarjan(int u,int pre)
{int v;Low[u] = DFN[u] = ++Index;Stack[top++] = u;Instack[u] = true;int son = 0;for(int i = head[u];i != -1;i = edge[i].next){v = edge[i].to;if(v == pre)continue;if( !DFN[v] ){son++;Tarjan(v,u);if(Low[u] > Low[v])Low[u] = Low[v];//桥//一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。if(Low[v] > DFN[u]){bridge++;edge[i].cut = true;edge[i^1].cut = true;}//割点//一个顶点u是割点，当且仅当满足(1)或(2) (1) u为树根，且u有多于一个子树。//(2) u不为树根，且满足存在(u,v)为树枝边(或称父子边，//即u为v在搜索树中的父亲)，使得DFS(u)<=Low(v)if(u != pre && Low[v] >= DFN[u])//不是树根
            {cut[u] = true;add_block[u]++;}}else if( Low[u] > DFN[v])Low[u] = DFN[v];}//树根，分支数大于1if(u == pre && son > 1)cut[u] = true;if(u == pre)add_block[u] = son - 1;Instack[u] = false;top--;
}void solve(int N)
{memset(DFN,0,sizeof(DFN));memset(Instack,false,sizeof(Instack));memset(add_block,0,sizeof(add_block));memset(cut,false,sizeof(cut));Index = top = 0;bridge = 0;for(int i = 1;i <= N;i++)if( !DFN[i] )Tarjan(i,i);printf("%d critical links\n",bridge);vector<pair<int,int> >ans;for(int u = 1;u <= N;u++)for(int i = head[u];i != -1;i = edge[i].next)if(edge[i].cut && edge[i].to > u){ans.push_back(make_pair(u,edge[i].to));}sort(ans.begin(),ans.end());//按顺序输出桥for(int i = 0;i < ans.size();i++)printf("%d - %d\n",ans[i].first-1,ans[i].second-1);printf("\n");
}
void init()
{tot = 0;memset(head,-1,sizeof(head));
}
//处理重边
map<int,int>mapit;
inline bool isHash(int u,int v)
{if(mapit[u*MAXN+v])return true;if(mapit[v*MAXN+u])return true;mapit[u*MAXN+v] = mapit[v*MAXN+u] = 1;return false;
}
int main()
{int n;while(scanf("%d",&n) == 1){init();int u;int k;int v;//mapit.clear();for(int i = 1;i <= n;i++){scanf("%d (%d)",&u,&k);u++;//这样加边，要保证正边和反边是相邻的，建无向图while(k--){scanf("%d",&v);v++;if(v <= u)continue;//if(isHash(u,v))continue;
                addedge(u,v);addedge(v,u);}}solve(n);}return 0;
}