WA8
Please, explain, what is the reason? Is this just a problem of input/output or this is just such test case?

So, we can decompose like this any connected graph G = (V, E), |E| = 2k, k in Z+? This is lovely, this is nice (I proved it using induction, I'm proud of myself, yay!).

Could anyone help me?

In case anybody wants to see implementation Link is here http://paste.ubuntu.com/13082792/.

My mistake was usage of queue when the correct is using a stack. Only changed that, and then got AC 0.015 sec 480 KB. I'm so happy :D

Edited by author 05.08.2015 05:31

It's very simple problem because all connected graphs which containes even number of edges can be divided into pairs of edges, which includes one common vertex!!!(and all connected graphs with odd number of edges can't be divided in such way)

input

5 4
4 2
2 1
1 3
3 4

I think output should "1" but most of my colleagues' accepted programs print "0". Do I misunderstand the problem ?

CONGRATULATIONS TO THE AUTHOR!!!!!

who can help me? I WA on #7.Why?

here is my program:
var
  a:array[1..1000]of longint;
  b:array[1..1000,1..1000]of boolean;
  n,i,j,x,y,m:longint;
  c:boolean;
  function find(x,y:longint):boolean;
    var
      i:longint;
    begin
      if b[x,y] then exit(false);
      for i:=1 to m do
        if (b[a[i],x])and(b[a[i],y])then
          begin
            b[x,y]:=true;b[y,x]:=true;
            exit(false);
          end;
      exit(true);
    end;

begin
  while not seekeof do
    begin
      readln(x,y);
      b[x,x]:=true;b[y,y]:=true;
      b[x,y]:=true;b[y,x]:=true;

      c:=true;
      for j:=1 to m do
        if a[m]=x then begin c:=false;break;end;
      if c then
        begin
          inc(m);a[m]:=x;
          for i:=1 to m do
            if find(a[i],x) then
              begin
                writeln(0);halt;
              end;
        end;

      c:=true;
      for j:=1 to m do
        if a[m]=y then begin c:=false;break;end;
      if c then
        begin
          inc(m);a[m]:=y;
          for i:=1 to m do
            if find(a[i],y) then
              begin
                writeln(0);halt;
              end;
        end;
    end;
  writeln(1);
end.

1 2
2 3
3 1
1 10
in this test , i do not know why ?
what about the others vertex ? 4,5,6,7,8,9 ??

begin
  randomize;
  writeln(random(2));
end.

Who can AC by this program???

Edited by author 25.12.2008 02:46

the subject used ufs?

Could anyone help me?

in pascal it is 'seekeof()', and in C++ ?

Could anyone help me?

Is test1 the sample?
I have WA#1 !!!

dsd

Can i write "1" if all connected components of graph have even number of edges?