This algorithm is simple, but timeout: var n,m,i,x,y:longint; f:array[1..1000,1..1000] of Boolean; Procedure Floyd; var k,i,j:longint; Begin for K:=1 to N does for I:=1 to N does for J:=1 to N do &N Bsp F[I,J]:=F[I,J] or (f[i,k) and f[k,j]); End Begin Fillchar (f,sizeof (f), false); Read (n,m); for I:=1 to M do begin read (x,y); F[x,y]:=true; END; Floyd; for I:=1 to n do if f[i,i] then Writeln (' T ') else write (' F '); End. The simple deep search still timed out: var n,m,a,b,i,d:longint; f:array[1..1000,1..1000] of Boolean; used:array[1..1000] of Boolean; Flag:boolean; Procedure Search (X:longint;var d:longint); var k:longint; Begin if (x=i) and (d>0) and (not flag) then begin writeln (' T '); Flag:=true; exit; &NBsp End for K:=1 to N does if (F[x,k]) and (not used[k]) then begin & nbsp Used[k]:=true; Inc. (d); search (k,d); Used[k]:=false; {Remove this sentence or timeout, only 8 points, do not backtrack} end; End Begin Fillchar (f,sizeof (f), false); Read (n,m); for I:=1 to M do begin read (A,B); F[a,b]:=true; END; for I:=1 to n do begin Fillchar (used,sizeof (used), false); Flag:=false; d:=0; Search (i,d); If not flag then Writeln (' F '); END; End. Weakly weak ground: Still use adjacent chain table to keep the edge ...
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