POJ 1836 Alignment

解题思路：正反方向最长递增子序列(参照POJ2533)，然后枚举中心

 

  
    

   
     
   
     #include 
   
     <
   
     iostream
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     int
   
      n;

   
     float
   
      num[
   
     1000
   
     ],B[
   
     1001
   
     ];
inline 
   
     void
   
      LIS(
   
     int
   
      
   
     *
   
     h, 
   
     bool
   
      isLeft)
{
 
   
     int
   
      i,j,p,q,m,len;
 
   
     if
   
     (
   
     !
   
     isLeft)
   
     for
   
     (i
   
     =
   
     0
   
     ,j
   
     =
   
     n
   
     /
   
     2
   
     ;i
   
     <
   
     j;i
   
     ++
   
     )swap(num[i],num[n
   
     -
   
     1
   
     -
   
     i]);
 
   
     for
   
     (B[
   
     0
   
     ]
   
     =
   
     0
   
     ,B[
   
     1
   
     ]
   
     =
   
     num[
   
     0
   
     ],h[
   
     0
   
     ]
   
     =
   
     len
   
     =
   
     1
   
     ,i
   
     =
   
     1
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
   
     //
   
     最长递增子序列
   
     

   
      {
 
   
     for
   
     (p
   
     =
   
     0
   
     ,q
   
     =
   
     len,m
   
     =
   
     (p
   
     +
   
     q)
   
     /
   
     2
   
     ;p
   
     <=
   
     q;m
   
     =
   
     (p
   
     +
   
     q)
   
     /
   
     2
   
     )
 
   
     if
   
     (B[m]
   
     <
   
     num[i])p
   
     =
   
     m
   
     +
   
     1
   
     ;
   
     else
   
      q
   
     =
   
     m
   
     -
   
     1
   
     ;
 B[p]
   
     =
   
     num[i];
   
     if
   
     (p
   
     >
   
     len)len
   
     ++
   
     ;h[i]
   
     =
   
     p;
 }
 
   
     if
   
     (
   
     !
   
     isLeft)
   
     for
   
     (i
   
     =
   
     0
   
     ,j
   
     =
   
     n
   
     /
   
     2
   
     ;i
   
     <
   
     j;i
   
     ++
   
     )swap(h[i],h[n
   
     -
   
     1
   
     -
   
     i]);
}

   
     int
   
      main()
{
 
   
     int
   
      i,j,k,l[
   
     1000
   
     ],r[
   
     1000
   
     ],ans,m1,m1i,m;
 scanf(
   
     "
   
     %d
   
     "
   
     , 
   
     &
   
     n);
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )scanf(
   
     "
   
     %f
   
     "
   
     , 
   
     &
   
     num[i]);
 LIS(l,
   
     1
   
     );LIS(r,
   
     0
   
     );
 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
   
     if
   
     (l[i]
   
     <
   
     l[i
   
     -
   
     1
   
     ])l[i]
   
     =
   
     l[i
   
     -
   
     1
   
     ];
 
   
     for
   
     (i
   
     =
   
     n
   
     -
   
     2
   
     ;i
   
     >=
   
     0
   
     ;i
   
     --
   
     )
   
     if
   
     (r[i]
   
     <
   
     r[i
   
     +
   
     1
   
     ])r[i]
   
     =
   
     r[i
   
     +
   
     1
   
     ];
 
   
     for
   
     (ans
   
     =
   
     i
   
     =
   
     0
   
     ;i
   
     <
   
     n
   
     -
   
     1
   
     ;i
   
     ++
   
     )
   
     if
   
     (ans
   
     <
   
     l[i]
   
     +
   
     r[i
   
     +
   
     1
   
     ])ans
   
     =
   
     l[i]
   
     +
   
     r[i
   
     +
   
     1
   
     ];//枚举中心点
 printf(
   
     "
   
     %d\n
   
     "
   
     , n
   
     -
   
     ans);
 
   
     return
   
      
   
     0
   
     ;
}