Usaco Training Section 3.3 shopping offers

Take each solution as an item
Other sold items are also used as an item.

Just take them to the backpack.

Pay attention to the corresponding number.

Code:

# Include <algorithm> # include <iostream> # include <cstring> # include <cstdlib> # include <cctype> # include <cstdio> # include <locale> # include <map> using namespace STD; const int maxn = 110; struct item {int tot, CST; int num [6];} ITM [maxn]; int S, N, totnum; int num [6], id [1005], req [1005], F [6] [6] [6] [6] [6] [6]; bool USF [1005]; int main () {scanf ("% d", & S); int x = 0, y = 0; For (INT I = 1; I <= s; ++ I) {scanf ("% d ", & ITM [I]. TOT); For (Int J = 1; j <= ITM [I]. TOT; ++ J) {scanf ("% d", & X, & Y); If (! Id [x]) {ID [x] = ++ totnum; num [totnum] = x;} ITM [I]. num [ID [x] = y;} scanf ("% d", & ITM [I]. CST);} scanf ("% d", & N); For (INT I = 1; I <= N; ++ I) {scanf ("% d ", & X); If (! Id [x]) {ID [x] = ++ totnum; num [totnum] = x;} scanf ("% d", & req [ID [x]); ITM [++ S]. num [ID [x] = 1; ITM [s]. TOT = 1; scanf ("% d", & ITM [s]. CST);} memset (F, 0x3f, sizeof (f); F [0] [0] [0] [0] [0] = 0; for (INT I = 1; I <= s; ++ I) {for (int A = ITM [I]. num [1]; A <= req [1]; ++ A) {for (int B = ITM [I]. num [2]; B <= req [2]; ++ B) {for (INT c = ITM [I]. num [3]; C <= req [3]; ++ c) {for (INT d = ITM [I]. num [4]; D <= req [4]; ++ d) {for (int e = ITM [I]. num [5]; e <= req [5]; ++ E) {f [a] [B] [C] [d] [e] = min (F [a] [B] [C] [d] [e], f [A-ITM [I]. num [1] [B-ITM [I]. num [2] [C-ITM [I]. num [3] [D-ITM [I]. num [4] [E-ITM [I]. num [5] + ITM [I]. CST) ;}}}} printf ("% d \ n ", f [req [1] [req [2] [req [3] [req [4] [req [5]); Return 0 ;}

Usaco Training Section 3.3 shopping offers