POJ-2754 similarity of necklaces 2 interval lower bound operation + dp

A clever operation for this question is to agree to take out the lower bound for a given range, and then convert it into a multi-backpack problem. Binary optimization is used.

CodeAs follows:

# Include <cstdlib> # Include <Cstring> # Include <Cstdio> # Include <Cmath> # Include <Algorithm> # Include <Iostream> # Define Er 0x80808080 Using   Namespace  STD;  /* Problem-solving: Given a multi array and a pairs array, the best table array is required. to meet these requirements, sum {table [I] * multi [I]} = 0 and the maximum solution of sum {table [I] * pairs} is required: calculate the lower bound as a part. Therefore, [L [I], U [I] is converted to [1, U [I]-l [I] multiple backpacks. both M [I] and P [I] perform separate lower bound calculations. calculate T = L [1] * m [1] + L [2] * m [2]... then there is a multi-backpack with just enough t capacity.  */  Int N, P [ 205 ], M [ 205 ], L [ 205 ], U [ 205  ];  Int Lim, DP [ 100005  ]; Void Zobag ( Int W, Int  P ){  For ( Int I = lim; I> = W ;-- I ){  If (DP [I-W]! = Er) DP [I] = Max (DP [I], DP [I-W] + P );}}  Int  DP (){  Int  Q, N, K; DP [ 0 ] = 0  ;  For ( Int I = 1 ; I <= N; ++ I ){ //  Lists the item numbers. K = 1  ;  While (U [I]-k> 0 ){ //  If so many pieces can be separated Zobag (K * m [I], K *P [I]); U [I] -= K; k <= 1  ;}  If  (U [I]) zobag (U [I] * M [I], U [I] * P [I]);}  Return  DP [Lim];}  Int  Main (){  Int  RET;  While (Scanf ("  % D  " , & N )! = EOF) {Lim = Ret = 0  ; Memset (DP,  0x80 , Sizeof  (DP ));  For ( Int I = 1 ; I <= N; ++ I) {scanf (  " % D  " , & P [I], & M [I], & L [I], & U [I]); Lim + = L [I] * m [I], U [I]-= L [I]; RET + = L [I] * P [I]; //  P [I] also needs to calculate the lower bound, and then add it back.  } Lim * =- 1 ; //  It must be a number not greater than 0, because it is already the minimum value, and the final result of the question is 0. Printf ( "  % D \ n " , Dp () + RET );}  Return   0  ;}