Hdu 5135 status compression dp

Hdu 5135 status compression dp

The question is to give the most 12 sides. You ask the maximum area of a triangle that can be formed (multiple triangles are allowed). Typical state compression (up to 12 sides)

Ideas:
Enumerate the area of all possible triangles in a brute force manner, and then compress the area based on the state to form up to n/3 triangles. Each triangle is recorded on three sides or (operation) merge with area and then determine (based on and calculate ).

#include
 
  #include
  
   #include
   
    #include
    
     using namespace std
     ;double dp
     [1
     <<14
     ];double max
     (double a
     ,double b
     ){    return a
     >b
     ?a
     :b
     ;}struct node
     {    int pp
     ;    double area
     ;}cont
     [1
     <<14
     ];int main(){    int n
     ,i
     ,j
     ,k
     ;    int num
     [15
     ];    double S
     ;    while(~scanf
     ("%d"
     ,&n
     ),n
     )    {        for(i
     =1
     ;i
     <=n
     ;i
     ++)        scanf
     ("%d"
     ,&num
     [i
     ]);        memset
     (dp
     ,0
     ,sizeof(dp
     ));        int t
     =0
     ;        S
     =0
     ;        for(i
     =1
     ;i
     <=n
     ;i
     ++)        for(j
     =i
     +1
     ;j
     <=n
     ;j
     ++)        {            for(k
     =j
     +1
     ;k
     <=n
     ;k
     ++)            {                if(num
     [i
     ]+num
     [j
     ]>num
     [k
     ]&&num
     [i
     ]+num
     [k
     ]>num
     [j
     ]&&num
     [j
     ]+num
     [k
     ]>num
     [i
     ])                {                    double p
     =1.0
     *(num
     [i
     ]+num
     [j
     ]+num
     [k
     ])/2
     ;                    t
     ++;                    double s
     =sqrt
     (p
     *(p
     -num
     [i
     ])*(p
     -num
     [j
     ])*(p
     -num
     [k
     ]));                    cont
     [t
     ].pp
     =(1
     <<i
     )|(1
     <<j
     )|(1
     <<k
     );                    dp
     [(1
     <<i
     )|(1
     <<j
     )|(1
     <<k
     )]=cont
     [t
     ].area
     =s
     ;                }            }        }        for(i
     =1
     ;i
     <(1
     <<(n
     +1
     ));i
     ++)        {            for(j
     =1
     ;j
     <=t
     ;j
     ++)            {                if((i
     &cont
     [j
     ].pp
     )==0
     )                {                    dp
     [i
     |cont
     [j
     ].pp
     ]=max
     (dp
     [i
     |cont
     [j
     ].pp
     ],dp
     [i
     ]+cont
     [j
     ].area
     );                    if(dp
     [i
     |cont
     [j
     ].pp
     ]>S
     ) S
     =dp
     [i
     |cont
     [j
     ].pp
     ];                }            }        }        printf
     ("%.2lf\n"
     ,S
     );    }    return 0
     ;}