HDU 3033 I love sneakers! (01 backpack variant)

There are n brands, and each brand must have at least one.

01 backpack with one-dimensional, design status DP [k] [I] indicates the maximum value of the first K spent I, the status can be obtained by the former K-1 and the current brand, pay attention to the initialization and equation location (no aftereffect ).

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 1   /*  2   Author: zhaofa Fang  3   Lang: C ++  4   */  5 # Include <cstdio> 6 # Include <cstdlib> 7 # Include <sstream> 8 # Include <iostream> 9 # Include <cmath> 10 # Include <cstring> 11 # Include <algorithm> 12 # Include < String > 13 # Include <utility> 14 # Include <vector> 15 # Include <queue> 16 # Include <stack> 17 # Include <map> 18 # Include < Set > 19   Using   Namespace  STD;  20   21 Typedef Long   Long  Ll;  22   # Define Debug (x) cout <# x <':' <x <Endl 23   # Define Rep (I, n) for (INT I = 0; I <(n); I ++)24   # Define For (I, S, T) for (INT I = (s); I <= (t); I ++) 25   # Define Ford (I, S, T) for (INT I = (s); I> = (t); I --) 26   # Define PII pair <int, int> 27   # Define PB push_back 28   # Define MP make_pair 29   # Define FT first 30  # Define SD second 31   # Define Lowbit (x) (X & (-x )) 32   # Define INF (1 <30) 33   34   Int DP [ 11 ] [ 10005  ];  35 Vector < Int > P [ 11 ], V [11  ];  36   Int  Main ()  37   {  38       //  Freopen ("in", "r", stdin );  39       //  Freopen ("out", "W", stdout );  40       Int  N, m, K;  41      While (~ Scanf ( "  % D  " , & N, & M ,& K ))  42   {  43           Int  A, B, C;  44 Rep (I, 11  ) P [I]. Clear (), V [I]. Clear ();  45   Rep (I, n)  46  {  47 Scanf ( "  % D  " , & A, & B ,& C );  48   P [A]. Pb (B); V [A]. Pb (C );  49   }  50 Memset (DP ,- 1 , Sizeof  (DP ));  51 Rep (I, m + 1 ) DP [ 0 ] [I] = 0  ;  52 For (K, 1  , K)  53   {  54   Rep (J, V [K]. Size ())  55   {  56   Ford (I, m, p [k] [J]) 57   {  58                       If (DP [k] [I-P [k] [J]! =- 1 ) //  If the two equations exchange positions, the non-aftereffect principle of DP is not met, which can be known from the second group of data.  59 DP [k] [I] = max (DP [k] [I], DP [k] [I-P [k] [J] + V [k] [J]);  60   61                       If (DP [k- 1 ] [I-P [k] [J]! =- 1 )  62 DP [k] [I] = max (DP [k] [I], DP [k- 1 ] [I-P [k] [J] + V [k] [J]);  63   64                       //  Debug (K), debug (I), debug (DP [k] [I]);  65   }  66   }  67   }  68          Int Ans =- 1  ;  69 For (J, 0 , M) ans = Max (ANS, DP [k] [J]);  70           If (ANS! =- 1 ) Printf ( "  % D \ n  "  , ANS );  71           Else Puts ("  Impossible  "  );  72   }  73       Return   0  ;  74   }  75   /*  76   3 5 3  77   1 2 5 78   2 2 1  79   3 2 2  80   81   3 5 3  82   1 0 5  83   2 0 1  84   3 0 2  85   86   5 10000 3  87   1 4 6 88   2 5 7  89   3 4 99  90   1 55 77  91   2 44 66  92   Ans:  93   Impossible  94   8  95   255  96   */