【POJ1144】Network（割点）（模板）

题意：给定一张无向图，求割点个数

思路：感谢CC大神http://ccenjoyyourlife.blog.163.com/的讲解

割点的定义就是某个联通块中删去此点连通性发生变化的的点

有两种割点：1.U为树根，子树个数>1

2.U非树根，有U的子节点V满足low[v]>=dfn[u]表示U的V子树必须通过U去到U的上面

更新时也有两种：dfn[u]<dfn[v]时u--->v 实边 反则u--->v 虚边

实边时low[u]=min(low[u],low[v]) 虚边low[u]=min(low[u],dfn[v])

 var head,vet,next,low,dfn,b,son,flag:array[..]of longint;
     n,m,x,i,tot,time,ans:longint;

 procedure add(a,b:longint);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  head[a]:=tot;
 end;

 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 procedure dfs(u:longint);
 var e,v:longint;
 begin
  flag[u]:=;
  inc(time); low[u]:=time; dfn[u]:=time;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    inc(son[u]);
    dfs(v);
    low[u]:=min(low[u],low[v]);
    if (low[v]>=dfn[u])and(u<>) then b[u]:=;
   end
    else low[u]:=min(low[u],dfn[v]);
   e:=next[e];
  end;
 end;

 begin
  assign(input,'1.in'); reset(input);
  assign(output,'1.out'); rewrite(output);
  repeat
   readln(n);
   if n= then break;
   fillchar(head,sizeof(head),);
   fillchar(low,sizeof(low),);
   fillchar(dfn,sizeof(dfn),);
   fillchar(flag,sizeof(flag),);
   fillchar(son,sizeof(son),);
   fillchar(b,sizeof(b),);
   repeat
    read(m);
    while not eoln do
    begin
     read(x);
     add(x,m);
     add(m,x);
    end;
   until m=;
   time:=;
   for i:= to n do
    if flag[i]= then dfs(i);
   ans:=;
   for i:= to n do if b[i]= then inc(ans);
   if son[]> then inc(ans);
   writeln(ans);
  until n=;
  close(input);
  close(output);
 end.