Poj 1014 Dividing // multiple backpacks

What are multiple backpacks?

There are N items and a backpack with a capacity of V. A maximum of Mi pieces are available for the I-th item. The space consumed by each item is Ci, and the value is Wi. Solving which items are loaded into a backpack makes the total space consumed by these items not exceed
Backpack capacity, and the total value is the largest.

Among them, if Mi * Ci> = V, then multiple backpacks are equivalent to a full backpack problem. Otherwise, convert multiple backpacks into 01 backpacks, but optimize them. (The optimization idea is clever ,).

For the pseudo code, see:

The code for this question is as follows:

/* Poj 1014 Dividing DP multiple backpack transfer equation: 01 The transfer equation of the backpack and the full backpack */# include <cstdio> # include <cstring> using namespace std; const int maxn = 120005; // do not enable const int N = 6; int f [maxn]; int a [N + 1]; int v; int totval; bool flag; int Max (int a, int B) {return a> B? A: B;} void ZeroOnePack (int c, int w) {for (int I = v; I> = c; I --) {f [I] = Max (f [I], f [I-c] + w); if (f [I] = v) flag = true ;}} void CompletePack (int c, int w) {for (int I = c; I <= v; I ++) {f [I] = Max (f [I], f [I-c] + w); if (f [I] = v) flag = true ;}} void MultiplePack (int c, int w, int m) {if (c * m> = v) {CompletePack (c, w); return;} int k = 1; while (k <m) {ZeroOnePack (k * c, k * w); m = m-k; k = k * 2;} Zer OOnePack (m * c, m * w);} int main () {int t = 0; while (scanf ("% d", & a [1], & a [2], & a [3], & a [4], & a [5], & a [6]) {t ++; totval = 0; for (int I = 1; I <= N; I ++) totval + = (I * a [I]); // printf ("totval = % d \ n", totval); if (totval = 0) break; if (totval % 2! = 0) {printf ("Collection # % d: \ n", t); printf ("Can't be divided. \ n "); continue;} memset (f,-1, sizeof (f); f [0] = 0; flag = false; v = totval/2; for (int I = 1; I <= N; I ++) {flag = false; MultiplePack (I, I, a [I]); if (flag = true) break;} if (flag) {printf ("Collection # % d: \ n", t); printf ("Can be divided. \ n ");} else {printf (" Collection # % d: \ n ", t); printf (" Can't be divided. \ n ") ;}} return 0 ;}

The code for multiple backpacks is very simple, but there are a few things to pay attention;

1) determine the transfer equation of the 01 backpack and the full backpack and the second cyclic order.

Ii). 01 the relationship between a backpack and multiple backpacks.

(3) Correctness of splitting.

4) f array initialization issues.

References: backpack 9 lecture