POJ 1159 Palindrome

解题思路：求解原串与其反串的LCS

  
    

   
     
   
     #include 
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     int
   
      main()
{
 
   
     int
   
      n,ai,bi,i,j,opt[
   
     2
   
     ][
   
     5001
   
     ]
   
     =
   
     {
   
     0
   
     };
 
   
     char
   
      a[
   
     5001
   
     ];
 scanf(
   
     "
   
     %d\n%s
   
     "
   
     ,
   
     &
   
     n, a);
 
   
     for
   
     (ai
   
     =
   
     0
   
     ,bi
   
     =
   
     i
   
     =
   
     1
   
     ;i
   
     <=
   
     n;i
   
     ++
   
     ,j
   
     =
   
     ai,ai
   
     =
   
     bi,bi
   
     =
   
     j)
 
   
     for
   
     (j
   
     =
   
     1
   
     ;j
   
     <=
   
     n;j
   
     ++
   
     )
 
   
     if
   
     (a[i
   
     -
   
     1
   
     ]
   
     ==
   
     a[n
   
     -
   
     j])opt[bi][j]
   
     =
   
     opt[ai][j
   
     -
   
     1
   
     ]
   
     +
   
     1
   
     ;
 
   
     else
   
      opt[bi][j]
   
     =
   
     max(opt[ai][j],opt[bi][j
   
     -
   
     1
   
     ]);
 printf(
   
     "
   
     %d\n
   
     "
   
     ,n
   
     -
   
     opt[ai][n]);
 
   
     return
   
      
   
     0
   
     ;
}