题解C++ | #购物单#

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int N, m;
    vector<vector<int>> pri(60, vector<int>(3, 0));
    vector<vector<int>> imp(60, vector<int>(3, 0));

    cin >> N >> m;
    for (int i = 1; i <= m; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        if (c == 0)
        {
            pri[i][0] = a;
            imp[i][0] = a * b;
        }
        else
        {
            if (pri[c][1] == 0)
            {
                pri[c][1] = a;
                imp[c][1] = a * b;
            }
            else
            {
                pri[c][2] = a;
                imp[c][2] = a * b;
            }
        }
    }
    vector<vector<int>> dp(m+1, vector<int>(N+1, 0));
    for (int i = 1; i <= m; i++)
    {
        for (int j = 1; j <= N; j++)
        {
            int a0 = pri[i][0], b0 = imp[i][0];
            int a1 = pri[i][1], b1 = imp[i][1];
            int a2 = pri[i][2], b2 = imp[i][2];
            // 当取主件时，和i-1件比
            dp[i][j] = j >= a0 ? max(b0+dp[i-1][j-a0], dp[i-1][j]) : dp[i-1][j];
            // 当取副件1时，和i-1件比，且i-1件中含有主件，如何体现含有主件，j减少了主件的价格，总体加了注件的满足感
            dp[i][j] = j >= (a0+a1) ? max(b0+b1+dp[i-1][j-a0-a1], dp[i][j]) : dp[i][j];
            // 取副件2同上
            dp[i][j] = j >= (a0+a2) ? max(b0+b2+dp[i-1][j-a0-a2], dp[i][j]) : dp[i][j];
            // 取副件1，2同上
            dp[i][j] = j >= (a0+a1+a2) ? max(b0+b1+b2+dp[i-1][j-a0-a1-a2], dp[i][j]) : dp[i][j];
        }
    }
    cout << dp[m][N];
}
// 64 位输出请用 printf("%lld")