Greedy solution for the longest ascending (descending) subsequence

The dynamic planning method is used to calculate the longest incrementing subsequence from the start to the end. We need to open two arrays.

 

Code

  # Include  <  Stdio. h  >  
  Int  Main ()
{
  Int  I, j, t, n, m;
  Int  A [  905 ], B [  905  ];
Scanf (  "  % D  "  ,  &  T );
  While  (T  --  ){
Scanf (  "  % D  "  , &  N,  &  M );
  For  (I  =  0  ; I  <  N; I  ++  )
Scanf (  "  % D  "  ,  & A [I]), B [I]  =  1  ;
  For  (I  =  1  ; I  <  N; I  ++  ){
  For  (J  =  0  ; J <  I; j  ++  ){
  If  (A [J]  <  A [I]  &&  B [J]  +  1  >  B [I])
B [I]  =  B [J]  + 1  ;
}
}
  //  Printf ("% d \ n", B [n-1]);  
    If  (B [n  -  1  ]  > =  M)
Puts (  "  Good  "  );
 Else  
Puts (  "  Bad  "  );
}
  Return     0  ;
} 

 

Until last night, the team members discussed with me, and I found that the greedy algorithm is used to find the longest incrementing subsequence. the space and time complexity are smaller than the dynamic planning method.

Method: J = 0; I starts from 0 to n-1, and each time we go to a [I]> A [J], let a [J + 1] = A [I], J + = 1; if a [I] <= A [J], replace the first number smaller than a [I] from a [I-1] in a [0] with a [I].

By the end of the loop, the value of J ++ is the result of obtaining the longest ascending subsequence.

Code

  # Include  <  Stdio. h  >  
  Int  Main ()
{
  Int  I, J, K, t, n, m;
  Int  A [  905  ];
Scanf (  " % D  "  ,  &  T );
  While  (T  --  ){
Scanf (  "  % D  "  ,  &  N,  &  M );
  For (I  =  0  ; I  <  N; I  ++  )
Scanf (  "  % D  "  ,  &  A [I]);
J  =  0  ;
  For (I  =  1  ; I  <  N; I  ++  ){
  If  (A [I]  >  A [J])
A [  ++  J]  =  A [I];
  Else 
  For  (K  =  0  ; K  <  I; k  ++  )
  If  (A [k]  > =  A [I]) {
A [k]  =  A [I];
  Break ;
}
}
  If  (  ++  J  > =  M)
Puts (  "  Good  "  );
  Else  
Puts (  "  Bad  " );
}
  Return     0  ;
} 

 

AboveCodeThe corresponding topic is the high jump snail on zzuli.