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])
1 var head,vet,next,low,dfn,b,son,flag:array[1..50000]of longint; 2 n,m,x,i,tot,time,ans:longint; 3 4 procedure add(a,b:longint); 5 begin 6 inc(tot); 7 next[tot]:=head[a]; 8 vet[tot]:=b; 9 head[a]:=tot; 10 end; 11 12 function min(x,y:longint):longint; 13 begin 14 if x<y then exit(x); 15 exit(y); 16 end; 17 18 procedure dfs(u:longint); 19 var e,v:longint; 20 begin 21 flag[u]:=1; 22 inc(time); low[u]:=time; dfn[u]:=time; 23 e:=head[u]; 24 while e<>0 do 25 begin 26 v:=vet[e]; 27 if flag[v]=0 then 28 begin 29 inc(son[u]); 30 dfs(v); 31 low[u]:=min(low[u],low[v]); 32 if (low[v]>=dfn[u])and(u<>1) then b[u]:=1; 33 end 34 else low[u]:=min(low[u],dfn[v]); 35 e:=next[e]; 36 end; 37 end; 38 39 begin 40 assign(input,'1.in'); reset(input); 41 assign(output,'1.out'); rewrite(output); 42 repeat 43 readln(n); 44 if n=0 then break; 45 fillchar(head,sizeof(head),0); 46 fillchar(low,sizeof(low),0); 47 fillchar(dfn,sizeof(dfn),0); 48 fillchar(flag,sizeof(flag),0); 49 fillchar(son,sizeof(son),0); 50 fillchar(b,sizeof(b),0); 51 repeat 52 read(m); 53 while not eoln do 54 begin 55 read(x); 56 add(x,m); 57 add(m,x); 58 end; 59 until m=0; 60 time:=0; 61 for i:=1 to n do 62 if flag[i]=0 then dfs(i); 63 ans:=0; 64 for i:=2 to n do if b[i]=1 then inc(ans); 65 if son[1]>1 then inc(ans); 66 writeln(ans); 67 until n=0; 68 close(input); 69 close(output); 70 end.