poj3177

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

 type node=record
        next,po:longint;
      end;

 var e:array[..] of node;
     p,dfn,low,d,be,fa:array[..] of longint;
     hash:array[..,..] of boolean;
     b:array[..] of boolean;
     len,n,m,x,y,i,ans,t,s,j,h,r:longint;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 procedure add(x,y:longint);
   begin
     hash[x,y]:=true;
     inc(len);
     e[len].po:=y;
     e[len].next:=p[x];
     p[x]:=len;
   end;

 procedure tarjan(x:longint);
   var i,y:longint;
   begin
     inc(h);
     dfn[x]:=h;
     low[x]:=h;
     i:=p[x];
     while i<>- do
     begin
       y:=e[i].po;
       if y<>fa[x] then
       begin
         if dfn[y]= then
         begin
           fa[y]:=x;
           tarjan(y);
         end;
         low[x]:=min(low[x],low[y]);
         if low[y]>dfn[x] then
         begin
           b[i]:=true;
           b[i xor ]:=true;
         end;
       end;
       i:=e[i].next;
     end;
   end;

 procedure dfs(x:longint);
   var i,y:longint;
   begin
     be[x]:=s;
     i:=p[x];
     while i<>- do
     begin
       y:=e[i].po;
       if (be[y]=) and not b[i] then dfs(y);
       if b[i] and (be[x]<>be[y]) then inc(d[be[x]]);
       i:=e[i].next;
     end;
   end;

 begin
   len:=-;
   fillchar(p,sizeof(p),);
   readln(n,m);
   for i:= to m do
   begin
     readln(x,y);
     if not hash[x,y] then
     begin
       add(x,y);
       add(y,x);
     end;
   end;
   tarjan();
   for i:= to n do
     if be[i]= then
     begin
       inc(s);
       dfs(i);
     end;

   for i:= to s do
     if d[i]= then inc(ans);
   writeln((ans+) shr );
 end.