​​5. 多重背包问题 II - AcWing题库​​​

输入样例

4 5
1 2 3
2 4 1
3 4 3
4 5 2

输出样例：

10

 朴素做法:

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

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

int main()
{
    cin >> N >> V;
    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 <= V; j ++ )
            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][V];
    return 0;
}

二维优化

将同类的物品分为若干堆,每一对看成新创造的新物品,之后按照01背包的思路即可

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 12010, M = 2010;

int v[N],w[N],f[M];

int main()
{
    int n,m;
    int a,b,s;
    int cnt = 0;
    cin >> n >> m;
    for (int i = 0; i < n; i ++ )
    {
        
        int k=1;
        cin >> a >> b >> s;
        while(k <= s){
            cnt++;
            v[cnt] = k*a;
            w[cnt] = k*b;
            s -= k;
            k *= 2;
        }
        if(s > 0){
            cnt++;
            v[cnt] = s*a;
            w[cnt] = s*b;
        }
    }
    
    for (int i = 1; i <= cnt; i ++ )
        for (int j = m; j >= v[i]; j -- )
            f[j] = max(f[j], f[j-v[i]]+w[i]);
            
    cout << f[m];
            
    return 0;
}