[ACWing] 4. Multiple knapsack problem I

Subject address:

https://www.acwing.com/problem/content/4/

Has $N$ items and one capacity are$V$ 's backpack. The first$i$ items have at most$s_i$Pieces, the volume of each piece is $v_i$, The value is $w_i$. Solve which items are loaded into the backpack, so that the total volume of the items does not exceed the capacity of the backpack, and the total value is the largest. Output the maximum value.

Data range:
$0<N,V\le 100$
$0<v_i,w_i,s_i\le 100$

The idea is dynamic programming. Let $f\left[i\right]\left[j\right]$ is if only the front$i$ items, and the total weight does not exceedjjThe maximum value that can be achieved in the case of$j$ . Then all schemes can follow theiiHow many$i$ items to take to classify. Obviously you can take at least$0$ , max can be$\min \{s_i,j/v_i\}$个，所以有 f [ i ] [ j ] = max ⁡ k = 0 , 1 , . . . , min ⁡ { s i , j / v i } { f [ i − 1 ] [ j − k v i ] + k w i } f[i][j]=\max_{k=0,1,...,\min\{s_i,j/v_i\}}\{f[i-1][j-kv_i]+kw_i\} f[i][j]=k = 0 , 1 , . . . , min { si​,j/vi​}max​{ f[i−1][j−kvi​]+kwi​} Finally return$f\left[N\right][V]\; is$ fine (there is a binary optimization for this question, refer tohttps://blog.csdn.net/qq_46105170/article/details/113840962). code show as below:

#include <iostream>
using namespace std;

const int N = 110;

int n, m;
int v[N], w[N], s[N];
int f[N][N];

int main() {
    
    
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> v[i] >> w[i] >> s[i];

    for (int i = 1; i <= n; i++) 
        for (int j = 0; j <= m; j++)
        	// 枚举第i件物品取多少个 
            for (int k = 0; k <= s[i] && k * v[i] <= j; k++)
                f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + k * w[i]);
    

    cout << f[n][m] << endl;

    return 0;
}

Time and space complexity $O\left(N{V}^2\right)$, space$O\left(NV\right)$。