【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[1..50000]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]:=1;
 inc(time); low[u]:=time; dfn[u]:=time;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   inc(son[u]);
   dfs(v);
   low[u]:=min(low[u],low[v]);
   if (low[v]>=dfn[u])and(u<>1) then b[u]:=1;
  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=0 then break;
  fillchar(head,sizeof(head),0);
  fillchar(low,sizeof(low),0);
  fillchar(dfn,sizeof(dfn),0);
  fillchar(flag,sizeof(flag),0);
  fillchar(son,sizeof(son),0);
  fillchar(b,sizeof(b),0);
  repeat
   read(m);
   while not eoln do
   begin
    read(x);
    add(x,m);
    add(m,x);
   end;
  until m=0;
  time:=0;
  for i:=1 to n do
   if flag[i]=0 then dfs(i);
  ans:=0;
  for i:=2 to n do if b[i]=1 then inc(ans);
  if son[1]>1 then inc(ans);
  writeln(ans);
 until n=0;
 close(input);
 close(output);
end.