【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])

 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.