long long findMaxProfit(Job Jobs[], long long n)
{
// Sort Jobs according to finish time
qsort(Jobs,n,sizeof(Job),cmpfunc);
long long table[100000 + 10];
table[0] = 1;
int i;
for (i=1; i < n; i++)
{
int inclProf = 1;
int l = latestNonConflict(Jobs, i);
if (l != -1)
inclProf += table[l];
if(inclProf > table[i-1])
table[i] = inclProf;
else
table[i] = table[i-1];
}
long long result = table[n-1];
return result;
}
//O(n^2)
int findMaxProfit(Job arr[], int n) {
// Sort jobs according to finish time
sort(arr, arr+n, myfunction);
// Create an array to store solutions of subproblems. table[i]
// stores the profit for jobs till arr[i] (including arr[i])
int *table = new int[n];
table[0] = arr[0].profit;
// Fill entries in M[] using recursive property
for (int i=1; i<n; i++) {
// Find profit including the current job
int inclProf = arr[i].profit;
int l = latestNonConflict(arr, i);
if (l != -1)
inclProf += table[l];
// Store maximum of including and excluding
table[i] = max(inclProf, table[i-1]);
}
int result = table[n-1];
delete[] table;
return result;
}
