边双连通有一个非常简单的做法就是先找出所有桥，然后再dfs一次不走桥即可

答案是（叶子节点的个数+1)/2

1 type node=record 2        next,po:longint; 3      end; 4  5 var e:array[0..20010] of node; 6     p,dfn,low,d,be,fa:array[0..5000] of longint; 7     hash:array[0..5000,0..5000] of boolean; 8     b:array[0..20010] of boolean; 9     len,n,m,x,y,i,ans,t,s,j,h,r:longint;10 11 function min(a,b:longint):longint;12   begin13     if a>b then exit(b) else exit(a);14   end;15 16 procedure add(x,y:longint);17   begin18     hash[x,y]:=true;19     inc(len);20     e[len].po:=y;21     e[len].next:=p[x];22     p[x]:=len;23   end;24 25 procedure tarjan(x:longint);26   var i,y:longint;27   begin28     inc(h);29     dfn[x]:=h;30     low[x]:=h;31     i:=p[x];32     while i<>-1 do33     begin34       y:=e[i].po;35       if y<>fa[x] then36       begin37         if dfn[y]=0 then38         begin39           fa[y]:=x;40           tarjan(y);41         end;42         low[x]:=min(low[x],low[y]);43         if low[y]>dfn[x] then44         begin45           b[i]:=true;46           b[i xor 1]:=true;47         end;48       end;49       i:=e[i].next;50     end;51   end;52 53 procedure dfs(x:longint);54   var i,y:longint;55   begin56     be[x]:=s;57     i:=p[x];58     while i<>-1 do59     begin60       y:=e[i].po;61       if (be[y]=0) and not b[i] then dfs(y);62       if b[i] and (be[x]<>be[y]) then inc(d[be[x]]);63       i:=e[i].next;64     end;65   end;66 67 begin68   len:=-1;69   fillchar(p,sizeof(p),255);70   readln(n,m);71   for i:=1 to m do72   begin73     readln(x,y);74     if not hash[x,y] then75     begin76       add(x,y);77       add(y,x);78     end;79   end;80   tarjan(1);81   for i:=1 to n do82     if be[i]=0 then83     begin84       inc(s);85       dfs(i);86     end;87 88   for i:=1 to s do89     if d[i]=1 then inc(ans);90   writeln((ans+1) shr 1);91 end.