Given n tasks with arrival time, priority and number of time units they need. We need to schedule these tasks on a single resource. The objective is to arrange tasks such that maximum priority tasks are taken. Objective is to minimize sum of product of priority and left time of tasks that are not scheduled due to limited time. This criteria simply means that the scheduling should cause minimum possible loss.

Examples:

Input : Total time = 3 Task1: arrival = 1, units = 2, priority = 300 Task2: arrival = 2, units = 2, priority = 100 Output : 100 Explanation : Two tasks are given and time to finish them is 3 units. First task arrives at time 1 and it needs 2 units. Since no other task is running at that time, we take it. 1 unit of task 1 is over. Now task 2 arrives. The priority of task 1 is more than task 2 (300 > 100) thus 1 more unit of task 1 is employed. Now 1 unit of time is left and we have 2 units of task 2 to be done. So simply 1 unit of task 2 is done and the answer is ( units of task 2 left X priority of task 2 ) = 1 X 100 = 100 Input : Total Time = 3 Task1: arrival = 1, units = 1, priority = 100 Task2: arrival = 2, units = 2, priority = 300 Output : 0

We use a priority queue and schedule one unit of task at a time.

- Initialize total loss as sum of each priority and units. The idea is to initialize result as maximum, then one by one subtract priorities of tasks that are scheduled.
- Sort all tasks according to arrival time.
- Process through each unit of time from 1 to total time. For current time, pick the highest priority task that is available. Schedule 1 unit of this task and subtract its priority from total loss.

Below is C++ implementation of above steps.

// C++ code for task scheduling on basis // of priority and arrival time #include "bits/stdc++.h" using namespace std; // t is total time. n is number of tasks. // arriv[] stores arrival times. units[] // stores units of time required. prior[] // stores priorities. int minLoss(int n, int t, int arriv[], int units[], int prior[]) { // Calculating maximum possible answer // that could be calculated. Later we // will subtract the tasks from it // accordingly. int ans = 0; for (int i = 0; i < n; i++) ans += prior[i] * units[i]; // Pushing tasks in a vector so that they // could be sorted according with their // arrival time vector<pair<int, int> > v; for (int i = 0; i < n; i++) v.push_back({ arriv[i], i }); // Sorting tasks in vector identified by // index and arrival time with respect // to their arrival time sort(v.begin(), v.end()); // Priority queue to hold tasks to be // scheduled. priority_queue<pair<int, int> > pq; // Consider one unit of time at a time. int ptr = 0; // index in sorted vector for (int i = 1; i <= t; i++) { // Push all tasks that have arrived at // this time. Note that the tasks that // arrived earlier and could not be scheduled // are already in pq. while (ptr < n and v[ptr].x == i) { pq.push({ prior[v[ptr].y], units[v[ptr].y] }); ptr++; } // Remove top priority task to be scheduled // at time i. if (!pq.empty()) { auto tmp = pq.top(); pq.pop(); // Removing 1 unit of task // from answer as we could // schedule it. ans -= tmp.x; tmp.y--; if (tmp.y) pq.push(tmp); } } return ans; } // driver code int main() { int n = 2, t = 3; int arriv[] = { 1, 2 }; int units[] = { 2, 2 }; int prior[] = { 100, 300 }; printf("%d\n", minLoss(n, t, arriv, units, prior)); return 0; }

Output:

100

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