k维背包

题目链接：E - Product Development (atcoder.jp)

因为最多为5，因此可以暴力枚举

int dp[10][10][10][10][10];
int a[110][10];
int n, k, p;
signed main()
{
    for (int i1 = 0; i1 <= 5; i1++)
        for (int i2 = 0; i2 <= 5; i2++)
            for (int i3 = 0; i3 <= 5; i3++)
                for (int i4 = 0; i4 <= 5; i4++)
                    for (int i5 = 0; i5 <= 5; i5++)
                        dp[i1][i2][i3][i4][i5] = 1145141919810;
    dp[0][0][0][0][0] = 0;
    cin >> n >> k >> p;
    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        for (int j = 1; j <= k; j++)
        {
            cin >> a[i][j];
        }
        for (int j = k + 1; j <= 5; j++)
        {
            a[i][j] = p;
        }
        for (int i1 = p; i1 >=0 ; i1--)
            for (int i2 = p; i2 >=0 ; i2--)
                for (int i3 = p; i3 >=0 ; i3--)
                    for (int i4 = p; i4 >=0 ; i4--)
                        for (int i5 = p; i5 >=0 ; i5--)
                        {
                            dp[min(p, i1 + a[i][1])][min(p, i2 + a[i][2])][min(p, i3 + a[i][3])][min(p, i4 + a[i][4])][min(p, i5 + a[i][5])] = min(dp[min(p, i1 + a[i][1])][min(p, i2 + a[i][2])][min(p, i3 + a[i][3])][min(p, i4 + a[i][4])][min(p, i5 + a[i][5])], dp[i1][i2][i3][i4][i5] + x);
                        }
    }
    if (dp[p][p][p][p][p] == 1145141919810)
        cout << -1;
    else
        cout << dp[p][p][p][p][p];
}

也可以使用动态规划。没太看懂应该是状态压缩吧（jiangly)的代码

int main()
{
    IOS;
    int n, k, p; cin >> n >> k >> p;
    vector<int> pw(k + 1);
    pw[0] = 1;
    for(int i = 1; i <= k; i ++)
        pw[i] = pw[i - 1] * (p + 1);
    vector<ll> dp(pw[k], -1ll);
    dp[0] = 0;
    for(int i = 0; i < n; i ++)
    {
        int c; cin >> c;
        vector<int> a(k);
        for(int j = 0; j < k; j ++) cin >> a[j];
        for(int s = pw[k] - 1; s >= 0; s --)
        {
            int t = 0;
            for(int j = 0; j < k; j ++)
            {
                int aa = s / pw[j] % (p + 1);
                int na = min(p, aa + a[j]);
                t += na * pw[j];
            }
            if(dp[s] != -1 && (dp[t] == -1 || dp[t] > dp[s] + c))
            {
                dp[t] = dp[s] + c;
            }
        }
    }
    cout << dp.back() << endl;
    return 0;
}