金明得预算方案(低配版有依赖的背包)

金明得预算方案(低配版有依赖的背包)

思路
样例分析：
1.该题目的依赖存储方式
2.用二进制方式构造分组背包
3.将有依赖的背包问题转化为分组背包从而成为0-1背包问题

样例输入：

1000 5
800 2 0
400 5 1
300 5 1
400 3 0
500 2 0

样例输出：

2200

代码模板：

#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>

#define v first
#define w second

using namespace std;

typedef pair<int, int> PII;

const int N = 60, M = 32010;

int n, m;
PII master[N];
vector<PII> servent[N];
int f[M];

int main()
{
    cin >> m >> n;

    for (int i = 1; i <= n; i ++ )
    {
        int v, p, q;
        cin >> v >> p >> q;
        p *= v;
        if (!q) master[i] = {v, p};
        else servent[q].push_back({v, p});
    }

    for (int i = 1; i <= n; i ++ )
        for (int u = m; u >= 0; u -- )
        {
            for (int j = 0; j < 1 << servent[i].size(); j ++ )
            {
                int v = master[i].v, w = master[i].w;
                for (int k = 0; k < servent[i].size(); k ++ )
                    if (j >> k & 1)
                    {
                        v += servent[i][k].v;
                        w += servent[i][k].w;
                    }
                if (u >= v) f[u] = max(f[u], f[u - v] + w);
            }
    }

    cout << f[m] << endl;

    return 0;
}

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