// https://www.cnblogs.com/stxy-ferryman/p/7779347.html
const int N=1005;
struct EDGE { int to, nt; }e[N*N];
int head[N], tot;
void addE(int u,int v) {
e[tot].to=v;
e[tot].nt=head[u];
head[u]=tot++;
}
int dfn[N], low[N], ind;
int col[N], id;
bool vis[N];
stack <int> s;
void init() {
while(!s.empty()) s.pop();
for(int i=0;i<=n;i++) {
head[i]=dfn[i]=low[i]=col[i]=-1;
vis[i]=du[i]=0;
}
tot=ind=id=0;
}
// 求强联通分量 并将各个分量染色为1~id
void tarjan(int u) {
dfn[u]=low[u]=ind++;
s.push(u); vis[u]=1;
for(int i=head[u];i!=-1;i=e[i].nt) {
int v=e[i].to;
if(dfn[v]==-1) {
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(vis[v])
low[u]=min(low[u],low[v]);
}
if(dfn[u]==low[u]) {
col[u]=++id;
vis[u]=0;
while(s.top()!=u) {
col[s.top()]=id;
vis[s.top()]=0;
s.pop();
} s.pop();
}
}
// 求割点 若i是割点 iscut[i]=1
int tarjanN(int u,int fa) {
int son=0, lowu;
dfn[u]=lowu=ind++;
for(int i=head[u];i!=-1;i=e[i].nt) {
int v=e[i].to;
if(dfn[v]==-1) {
son++;
int lowv=tarjan(v,u);
lowu=min(lowu,lowv);
if(lowv>dfn[u]) iscut[u]=1;
}
else if(v!=fa && dfn[v]<dfn[u])
low[u]=min(low[u],low[v]);
}
if(fa<0 && son==1) iscut[u]=0;
return low[u]=lowu;
}
// 求割边 边的信息存入E
vector < pair<int,int> > E;
int tarjanE(int u,int fa) {
int lowu;
dfn[u]=lowu=ind++;
for(int i=head[u];i!=-1;i=e[i].nt) {
int v=e[i].to;
if(dfn[v]==-1) {
int lowv=tarjan(v,u);
lowu=min(lowu,lowv);
if(lowv>dfn[u]) {
int from=u, to=v;
if(from>to) swap(from,to);
E.push_back(make_pair(from,to));
}
}
else if(v!=fa && dfn[v]<dfn[u])
low[u]=min(low[u],low[v]);
}
return low[u]=lowu;
}