But i receive WA2.

Problem is standart transative closer algorithm using DFS
Interesting feater that in chaina football player may be stronger than himself because of circle.
Let P(a)=list of Ancestors and GrandAncestors and .. of a;
Then a,b-YES ~P(a)^P(b)=Empty
P(a) would better have as bitset
I have TLE 24 because forgotten reachabylity algorithm O(n)
For our forum shold maintain standard algorithm Use library please!

It seems that transative closure as a rule has o(n^3) therefore when n=1000 we have no chances. But by god help strong components can be calculated for linear time and we can use forest of strong components. No for players a and b if and only if a and be dosn't belong to the same tree in this fotest.

So what about sample? 2 and 3 belong to the same tree, but answer is "No".
I thought, answer is "No", only if a and b common parent and there is no way from a to b.
Am i right?

1. DFS for all parent nodes
2. For each node store its parent in matrix [N][N/8] (bit mask)
3. For each A and B - if there is a common parent return "No"

Excuse me!. I were mistaken.
But I am shure that answer can be taken in terms of forest of strong components or Gerz graf.
Let Sc(a)- number of component for a;
    NumT(i) - number of tree in the forest for comp i
    Ncomp[i]- amount of vertex in i-th comp
    Root[i]=1 if i- root in forest
Then:
1) if (Sc(a)==Sc(b)) and (a!=b)) F(a,b)=No;
2) if NumT(Sc(a))==NumT(Sc(b)) and Root(Sc(a))==0
   and Root(Sc(b))==0 F(a,b)=No;
3) if NumT(Sc(a))==NumT(Sc(b)) and Root(Sc(a))==1
   and Ncomop(Sc(a)>1) F(a,b)=No;
4) if NumT(Sc(a))==NumT(Sc(b)) and Root(Sc(b))==1
   and Ncomop(Sc(b)>1) F(a,b)=No;

Otherwise F(a,b)=YES
Algo for strong comps has O(|V|+|E|)=O(n^2)
But I have WA4
If your don't agree with my arguments send message ,plese!

I finished and have Ac (0.6) on my Idea. All is right but final refinement says that sftrong component i may belong to many trees and we must use SubT[1..Nco][1..NT] matrix instead of array SubT[1..Nco].I used my previous strong components program algo O(|V|+O(|E|) from Kormen where i used vector container for adjacent lists this is the reason of bad time of executing. This is interface of the program:
void S_Comps(vector<short>smeg[MAX],short Nv,short &Nco,bool korn[MAX], short Scomps[MAX],char *Sm,short &NT,char *SubT,short Ncomp[MAX])
After O(n^3) Floid difficulties program has passed all tests immediately after removing silly mistakes.

I doubt there are any strong components larger than 1 (otherwise words "stronger" and "beaten" in problem statement loose big part of their original meaning).

Also, strong components form DAG, not necessarily a forest.

My Floyd in O(N^3) got AC within 0.25 sec.