解题思路

朴素的多重背包，在朴素的完全背包的基础，加一个次数限制，即可

朴素做法：一个一个拆 $O(N)$

二进制优化：类似于快速幂的拆法 $O(logN)$，拆成多组

比如说：a 物品有 13 个

朴素做法就是一个一个的试，最终试到 13 个

二进制优化，则是拆成 5 组，{1, 2, 4, 8, 1} 尝试。因为由 {1, 2, 4, 8, 1} 这个集合里面的子集能凑 0 ~ 13 的任意一个数

代码

原始做法 O(N^3)

#include <iostream>

using namespace std;

const int N = 110;
int v[N], w[N], s[N], f[N][N];

int main() {
    int n, m;
    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 = 1; j <= m; j++) {
            for (int k = 0; j >= k * v[i] && k <= s[i]; 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;
}

二进制优化 O(N^2logN)

#include <iostream>

using namespace std;

const int N = 10000;
int v[N], w[N], f[N][N];

int main() {
    int n, m;
    cin >> n >> m;
    int cnt = 1;
    for (int i = 1; i <= n; i++) {
        int a, b, s;
        cin >> a >> b >> s;
        for (int k = 1; k <= s; s -= k, k <<= 1) {
            v[cnt] = a * k, w[cnt] = b * k;
            cnt++;
        }
        if (s) {
            v[cnt] = a * s, w[cnt] = b * s;
            cnt++;
        }
    }

    n = cnt;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= v[i]) f[i][j] = max(f[i][j], f[i - 1][j - v[i]] + w[i]);
        }
    }
    cout << f[n][m] << endl;

    return 0;
}

在二进制优化上再优化空间

#include <iostream>

using namespace std;

const int N = 10000;
int v[N], w[N], f[N];

int main() {
    int n, m;
    cin >> n >> m;
    int cnt = 1;
    for (int i = 1; i <= n; i++) {
        int a, b, s;
        cin >> a >> b >> s;
        for (int k = 1; k <= s; s -= k, k <<= 1) {
            v[cnt] = a * k, w[cnt] = b * k;
            cnt++;
        }
        if (s) {
            v[cnt] = a * s, w[cnt] = b * s;
            cnt++;
        }
    }

    n = cnt;
    for (int i = 1; i <= n; i++) {
        for (int j = m; j >= v[i]; j--) {
            f[j] = max(f[j], f[j - v[i]] + w[i]);
        }
    }
    cout << f[m] << endl;

    return 0;
}