To solve this task problem using dynamic programming approach, we consider these two conditions,

1. If we will choose the low effort task then we have to add up this value with the value at (i-1)th index of the calculating array.
2. If we will choose the high effort task then we have to add up this value with the value at (i-2)th index of the calculating array.

After getting the value we will choose the maximum value between them.

Example:

5
4 5 9 12 5
3 6 4 5 4

C++ Implementation:

#include <iostream>
using namespace std;

int max_amount(int* high, int* low, int n)
{
    int arr[n + 1] = { 0 };
    arr[1] = max(high[0], low[0]);

    for (int i = 2; i <= n; i++) {
        arr[i] = max(arr[i - 1] + low[i - 1], arr[i - 2] + high[i - 1]);
    }

    return arr[n];
}

int main()
{
    int t;

    cout << "Test Case : ";
    cin >> t;

    while (t--) {
        int n;

        cout << "Enter the number of elements : ";
        cin >> n;

        cout << "Enter the High values : ";
        int high[n], low[n];
        for (int i = 0; i < n; i++) {
            cin >> high[i];
        }

        cout << "Enter the low values : ";
        for (int i = 0; i < n; i++) {
            cin >> low[i];
        }

        cout << "Maximum values : " << max_amount(high, low, n) << endl;
    }

    return 0;
}

Output

Test Case : 2
Enter the number of elements : 5
Enter the High values : 4 5 9 12 5
Enter the low values : 3 6 4 5 4
Maximum values : 26
Enter the number of elements : 5
Enter the High values : 2 4 7 9 10
Enter the low values : 3 2 8 7 6
Maximum values : 26