Sgu 214 weird dissimilarity

Sgu_214

This topic is similar to the longest common subsequence DP process. You can use F [I] [J] to indicate the optimal solution when λ matches I and μ matches J. Then f [I] [J] will only be obtained in three cases: one case is that λ [I] matches a character (obviously it should match the character with the smallest value added after matching ), one case is that μ [J] matches a character (it should also match the character with the smallest value added after matching ), another case is that λ [I] matches μ [J.

# Include <stdio. h>
# Include < String . H>
 # Define Maxd 2010
 # Define INF 0x7fffffff
 Char A [maxd], B [maxd], Alph [ 300 ];
 Long  Long   Int F [maxd] [maxd];
 Int N, Na, Nb, Mina [maxd], minb [maxd], G [maxd] [maxd], p [maxd] [maxd];
 Int C [ 600 ], * Ch, ANSA [ 2 * Maxd], Pa, ansb [ 2 * Maxd], PB;
 Void Init ()
{
 Int I, J, K;
Ch = & C [ 300 ];
N = strlen (Alph + 1 );
Gets (a + 1 );
NA = strlen (a + 1 );
Gets (B + 1 );
NB = strlen (B + 1 );
 For (I = 1 ; I <= N; I ++)
Ch [Alph [I] = I;
 For (I = 1 ; I <= N; I ++)
{
K = inf;
 For (J = 1 ; J <= N; j ++)
{
Scanf ( "  % D  " , & G [I] [J]);
 If (G [I] [J] <= K)
K = G [I] [J], Mina [I] = J;
}
}
 For (J = 1 ; J <= N; j ++)
{
K = inf;
 For (I = 1 ; I <= N; I ++)
 If (G [I] [J] <= K)
K = G [I] [J], minb [J] = I;
}
}
 Void DFS ( Int Na, Int NB)
{
 If (NA = 0 & Nb =0 )
 Return ;
 If (P [Na] [Nb] = 0 )
{
ANSA [PA ++] = CH [A [Na], ansb [Pb ++] = CH [B [Nb];
DFS (Na- 1 , Nb- 1 );
}
 Else   If (P [Na] [Nb]> 0 )
{
ANSA [PA ++] = CH [A [Na], ansb [Pb ++] = P [Na] [Nb];
DFS (Na- 1 , Nb );
}
 Else 
{
ANSA [PA ++] =-P [Na] [Nb], ansb [Pb ++] = CH [B [Nb];
DFS (Na, Nb- 1 );
}
}
 Void Printresult ()
{
 Int I, J, K;
Pa = Pb = 0 ;
DFS (Na, Nb );
Printf ( "  % I64d \ n  " , F [Na] [Nb]);
 While (PA)
Printf ( "  % C  " , Alph [ANSA [-- PA]);
Printf ( "  \ N  " );
 While (PB)
Printf ("  % C  " , Alph [ansb [-- Pb]);
Printf ( "  \ N  " );
}
 Void Solve ()
{
 Int I, J, K, X, Y;
 Long   Long   Int T;
F [ 0 ] [0 ] = 0 ;
T = 0 ;
 For (I = 1 ; I <= Na; I ++)
{
X = CH [A [I];
F [I] [ 0 ] = F [I- 1 ] [ 0 ] + G [x] [Mina [X], p [I] [ 0 ] = Mina [x];
}
T =0 ;
 For (I = 1 ; I <= Nb; I ++)
{
Y = CH [B [I];
F [ 0 ] [I] = f [ 0 ] [I- 1 ] + G [minb [y] [Y], p [ 0 ] [I] =-minb [y];
}
 For (I = 1 ; I <= Na; I ++)
 For (J = 1 ; J <= Nb; j ++)
{
X = CH [A [I], y = CH [B [J];
F [I] [J] = f [I- 1 ] [J- 1 ] + G [x] [Y], p [I] [J] = 0 ;
 If (F [I- 1 ] [J] + G [x] [Mina [x] <F [I] [J])
F [I] [J] = f [I- 1 ] [J] + G [x] [Mina [X], p [I] [J] = Mina [x];
 If (F [I] [J-1 ] + G [minb [y] [Y] <F [I] [J])
F [I] [J] = f [I] [J- 1 ] + G [minb [y] [Y], p [I] [J] =-minb [y];
}
Printresult ();
}
 Int Main ()
{
Gets (Alph + 1 );
Init ();
Solve ();
 Return   0 ;
}