Poj 2983 is the information reliable?

Poj_2983

With the foundation of the previous questions, this question is quite well thought. We set S [I] Point I Distance to a reference point on a straight line (you can set 0 As a reference point), in order to write the question, simply put the question A north
B Imagine it all B South than , If it is input P There should be two constrained equations. S [B]-s [a]> = x , S [a]-s [B]> =-x And input V Only corresponds to one constrained equation. S [B]-s [a]> = 1 .

Then, if we use the shortest short circuit, we will actually judge whether there is a negative circle in this graph. To make the Graph Connected, we abstract a source point. 0 To any point I All have S [I]-s [0]> = 0 In this way, the graph is connected. 0 Start point. You can check whether there is a negative circle.

 # Include  <  Stdio. h  >  
# Include  <  String  . H  >  
  Int  First [  1010  ], Next [  210000 ], V [  210000  ], W [  210000  ];
  Int  D [  1010  ], Q [  1010  ], INQ [  1010  ], Inedq [  1010  ];
  Char  B [ 5  ];
  Int  Main ()
{
  Int  I, J, K, a, B, X, n, m, U, E, OK, N, front, rear;
  While  (Scanf (  "  % D  "  ,  &  N,  &  M)  = 2  )
{
E  =  0  ;
Memset (first,  -  1  ,  Sizeof  (First ));
  For  (I  =  0  ; I  <  M; I ++  )
{
Scanf (  "  % S  "  , B );
  If  (B [  0  ]  =  '  V  '  )
{
Scanf (  "  % D "  ,  &  A,  &  B );
V [E]  =  A;
W [E]  =-  1  ;
Next [E]  =  First [B];
First [B]  =  E;
E  ++ ;
}
  Else     If  (B [  0  ]  =  '  P  '  )
{
Scanf (  "  % D  "  ,  & A,  &  B,  &  X );
V [E]  =  A;
W [E]  =-  X;
Next [E]  =  First [B];
First [B]  =  E;
E  ++  ;
V [E] =  B;
W [E]  =  X;
Next [E]  =  First [a];
First [A]  =  E;
E  ++  ;
}
}
  For  (I  =  1  ; I  <= N; I  ++  )
{
V [E]  =  I;
W [E]  =  0  ;
Next [E]  =  First [  0  ];
First [  0  ]  =  E;
E ++  ;
}
  For  (I  =  0  ; I  <=  N; I  ++  )
{
D [I]  =  1000000000  ;
INQ [I]  =  Inedq [I]  = 0  ;
}
D [  0  ]  =  0  ;
Front  =  Rear  =  0  ;
Q [rear  ++  ]  =  0  ;
OK =  1  ;
N  =  N  +  1  ;
  While  (Front  ! =  Rear)
{
U  =  Q [Front  ++  ];
INQ [u]  = 0  ;
  If  (Front  >  N)
Front  =  0  ;
  For  (E  =  First [u]; e  ! =-  1  ; E  = Next [e])
  If  (D [u]  +  W [E]  <  D [V [e])
{
D [V [e]  =  D [u]  +  W [E];
  If  (  !  INQ [V [e])
{
Q [rear  ++ ]  =  V [E];
  If  (Rear  >  N)
Rear  =  0  ;
INQ [V [e]  =  1  ;
  If  (  ++  Inedq [V [e] >  N)
{
OK  =  0  ;
Front  =  Rear;
  Break  ;
}
}
}
}
  If  (OK)
Printf (  "  Reliable \ n  " );
  Else  
Printf (  "  Unreliable \ n  "  );
}
  Return     0  ;
}