describe
Wang Qiang is very happy today. The company gives a year-end bonus of N yuan. Wang Qiang decided to use the year-end bonus for shopping. He divided the items he wanted to buy into two categories: main parts and accessories. Accessories are subordinate to a main part. The following table is an example of some main parts and accessories:
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Main accessories
Computer printer, scanner
Bookcase books
Desk lamp, stationery
Work chair none
*
If you want to buy an item classified as an accessory, you must first buy the main part to which the accessory belongs. Each master can have 0, 1 or 2 accessories. The attachment no longer has its own attachment. Wang Qiang wants to buy a lot of things. In order not to exceed the budget, he sets an importance degree for each item, which is divided into five grades: expressed by integers 1 ~ 5, grade 5 is the most important. He also found the price of each item on the Internet (an integral multiple of 10 yuan). He hopes to maximize the sum of the product of the price and importance of each item on the premise that it does not exceed N yuan (which can be equal to N yuan).
Let the price of the j-th item be v[j] and its importance be w[j]. k items are selected, and the numbers are j 1, j 2,..., J k, then the sum is: v[j 1] * w[j 1] + v[j 2] * w[j 2] +... + v[j k] * w[j k]. (where * is the multiplication sign)
Please help Wang Qiang design a shopping list that meets the requirements.
*
Input Description: the first line of input is two positive integers separated by a space: N m (where N (< 32000) represents the total money and m (< 60) represents the number of items you want to buy.)
*
From line 2 to line m+1, line j gives the basic data of the article numbered j-1. Each line has three non negative integers v p q (where V represents the price of the article (V < 10000) and P represents the importance of the article (1 ~ 5) , Q indicates whether the item is a main part or an accessory. If q=0, it indicates that the item is a main part; if Q > 0, it indicates that the item is an accessory, and Q is the number of the main part)
*
Output Description: the output file has only one positive integer, which is the maximum value (< 200000) of the sum of the product of the price and importance of the goods that do not exceed the total money.
*
Example 1
Input:
1000 5
800 2 0
400 5 1
300 5 1
400 3 0
500 2 0
Output:
2200

Solution – dynamic planning

#include<iostream>
#include<vector>
using namespace std;
int max(int m, int n)
{
    return m>n?m:n;
}
int dp[3200];
int main()
{
    int N,n,v,p,q;
    cin >> N >> n;
    N = N/10;
 
    int *ZJ_Pri = new int[n+1]();  int *ZJ_Imp = new int[n+1]();
    int *FJ1_Pri = new int[n+1](); int *FJ1_Imp = new int[n+1]();
    int *FJ2_Pri = new int[n+1](); int *FJ2_Imp = new int[n+1]();
 
    for(int i=1; i<=n; i++)
    {
        cin >> v >> p >> q;
        v = v / 10;
        if(q == 0)
        {
            ZJ_Pri[i] = v;
            ZJ_Imp[i] = v * p;
        }
        else if(FJ1_Pri[q] == 0)
        {
            FJ1_Pri[q] = v;
            FJ1_Imp[q] = v * p;
        }
        else
        {
            FJ2_Pri[q] = v;
            FJ2_Imp[q] = v * p;
        }
    }
    for(int i = 1; i <= n; i++)//i --- first i items
    {
        for(int j = N; j >=1; j--)//j -- current amount of money
        {
 
            if(j >= ZJ_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i] ] + ZJ_Imp[i]);
            if(j >= ZJ_Pri[i] + FJ1_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i]-FJ1_Pri[i] ] + ZJ_Imp[i] + FJ1_Imp[i]);
            if(j >= ZJ_Pri[i] + FJ2_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i]-FJ2_Pri[i] ] + ZJ_Imp[i] + FJ2_Imp[i]);
            if(j >= ZJ_Pri[i] + FJ1_Pri[i] + FJ2_Pri[i])
                dp[j] = max(dp[j], dp[ j-ZJ_Pri[i]-FJ1_Pri[i]-FJ2_Pri[i] ] + ZJ_Imp[i] + FJ1_Imp[i] + FJ2_Imp[i]);
        }
    }
    cout << dp[N]*10 << endl;
 
    delete[] ZJ_Pri;
    delete[] ZJ_Imp;
    delete[] FJ1_Pri;
    delete[] FJ1_Imp;
    delete[] FJ2_Pri;
    delete[] FJ2_Imp;
 
    return 0;
}