I use DFS, but get WA.
Give me a test please.

Andrey Popyk.
popyk@ief.tup.km.ua
ICQ# 88914410

CONST Dim = 7200;

VAR A:Array[1..Dim,0..3] of integer;
    Col:Array[1..Dim] of byte;
    N:integer;

PROCEDURE ReadData;
var i,j:integer;
begin
  readln(N);
  for i:=1 to N do
  begin
    read(A[i,0]);
    for j:=1 to A[i,0] do read(A[i,j]);
    readln;
  end;
end;

PROCEDURE DFS(v:integer; c:byte);
var i:integer;
begin
  Col[v]:=c;
  for i:=1 to A[v,0] do
    if Col[A[v,i]]=0 then DFS(A[v,i],3-c);
end;

PROCEDURE Solve;
var i:integer;
begin
  for i:=1 to N do
    if Col[i]=0 then DFS(i,1);
end;

PROCEDURE WriteData;
var S,C,i:integer;
begin
  S:=0;
  for i:=1 to N do
    if Col[i]=1 then inc(S);
  if S>N-S then begin S:=N-S; C:=2 end
           else C:=1;
  writeln(S);
  for i:=1 to N do
    if Col[i]=C then write(i,' ');
  writeln;
end;

BEGIN
  ReadData;
  Solve;
  WriteData;
END.

How about this?

4
3 2 3 4
3 1 3 4
2 1 2
2 1 2

DFS does this:

Go to node 1, mark it as colour 1.
Go to node 2, mark it as colour 2.
Go to node 3, mark it as colour 1.
Return to node 2.
Go to node 4, mark it as colour 1.

Then node 1 will be the same colour as node 3 and 4.

Correct solution is to put nodes 3 and 4 together and node 1 and 2
together.

Did I make any error here?