CF 105059D - Assignment Allocation

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Solution

Problem Understanding

We are given a semester divided into days, and a collection of assignments. Each assignment is available only during a fixed interval of days, from its start day to its deadline day, and it can be completed on any single day inside that interval. However, you are extremely constrained in how much work you are willing to do: at most one assignment per day.

The goal is not to maximize completed assignments directly, but to minimize the number of assignments that you fail to submit. There is an additional twist: the professor allows you to “forgive” up to k missed assignments, meaning only the remaining missed assignments beyond k actually hurt your score.

So effectively, if you complete X assignments, then n − X is the number of missed assignments, and your final penalty is max(0, n − X − k). Since k is fixed, the core task is to maximize how many assignments you can complete under the constraint of one assignment per day, while respecting availability windows.

The structure is a classic scheduling problem: each assignment is a unit-length job with a time window, and each day can host at most one job.

The constraints are large: up to 2 × 10^5 assignments per test and total input size across tests also bounded by 2 × 10^5. This rules out anything quadratic such as checking each assignment against every day or repeatedly scanning active intervals. Any solution must be roughly O(n log n) or O(n) per test.

A subtle edge case appears when many assignments share the same narrow interval. For example, if two assignments both have interval [1,1], only one can be completed, and the other is inevitably lost even if k is large. Another corner case is when intervals are very wide but scattered; a greedy choice of “earliest start first” without considering deadlines can waste early days and block later feasible assignments.

Approaches

The brute-force idea is straightforward: simulate day by day, and on each day pick any available assignment that has not been completed yet. For each day, we scan all assignments to find those whose interval includes that day, and choose one. This is correct because it respects both constraints: at most one per day and only selecting valid assignments. However, for each of d days we may inspect up to n assignments, producing O(nd) behavior, which is far too slow when both n and d are large.

The key observation is that we do not actually need to think in terms of days as the primary axis. Instead, we can think of assignments as events that become “eligible” over time. Once we reach a day t, all assignments with Si ≤ t become candidates, and among those, we want to avoid wasting opportunities on assignments that expire soon.

This leads to a greedy scheduling strategy: process days in order, maintain a structure of all currently available assignments, and always pick the one with the earliest deadline. This is the classic “interval scheduling with release times and deadlines” under unit capacity per time step. The intuition is that choosing a long-deadline assignment early can block a tight-deadline assignment later, while the reverse is safe.

To avoid iterating over all days explicitly, we instead simulate only relevant days by sweeping through time and maintaining a priority structure of active intervals.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force simulation per day scanning all jobs	O(n·d)	O(n)	Too slow
Greedy with min-heap over deadlines	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

We treat each assignment as entering the system on its start day and leaving after its end day. We sweep through days in increasing order, but we only explicitly advance when events occur.

Sort all assignments by start day. This allows us to activate them in chronological order without repeatedly scanning the full list.
Maintain a pointer over the sorted assignments and a min-heap keyed by end day. The heap represents all assignments that are currently available but not yet completed.
Iterate through days from 1 to d. Before processing day t, insert into the heap all assignments whose start day is ≤ t. This ensures the heap always contains exactly the feasible assignments for day t.
Remove from the heap any assignments whose end day is < t, since they can no longer be completed. This step prevents invalid choices from lingering.
If the heap is non-empty, assign day t to the job with the smallest end day. This greedy choice ensures we consume the most urgent assignment first.
Continue until all days are processed or all assignments are exhausted. The number of completed assignments is the total number of successful heap pops used for assignment.
Convert completion count into penalty by computing n − completed, then subtract k forgiveness and clamp at zero.

Why it works

The heap always prioritizes the assignment that will expire first among all currently feasible ones. Suppose we ever pick a job with a later deadline while a tighter one exists. That tight job could still be scheduled only on a subset of remaining days, so skipping it risks permanent loss that cannot be repaired later. By always consuming the earliest-deadline job, we preserve flexibility for all others, which is exactly the greedy exchange argument that guarantees optimal scheduling for unit-capacity interval tasks.

Python Solution

import sys
input = sys.stdin.readline

import heapq

def solve():
    t = int(input())
    out = []

    for _ in range(t):
        n, d, k = map(int, input().split())
        jobs = []
        for _ in range(n):
            s, e = map(int, input().split())
            jobs.append((s, e))

        jobs.sort()
        i = 0
        heap = []
        done = 0

        for day in range(1, d + 1):
            while i < n and jobs[i][0] <= day:
                heapq.heappush(heap, jobs[i][1])
                i += 1

            while heap and heap[0] < day:
                heapq.heappop(heap)

            if heap:
                heapq.heappop(heap)
                done += 1

        missed = n - done
        out.append(str(max(0, missed - k)))

    print("\n".join(out))

if __name__ == "__main__":
    solve()

The code first sorts assignments by start day so that we can incrementally activate them as the sweep progresses. The heap stores only end days, since start constraints are already satisfied when inserted.

The cleanup loop removes expired assignments, ensuring correctness even when many jobs become invalid over time. The greedy pop always corresponds to scheduling a task on that day.

Finally, we compute the penalty using the forgiveness parameter k.

Worked Examples

Example 1

Input:

n=3, d=3, k=1
(1,2), (1,3), (2,3)

We track heap contents and decisions:

Day	Newly added	Heap after add	Removed expired	Chosen job	Done
1	(1,2),(1,3)	[2,3]	none	(1,2)	1
2	(2,3)	[3,3]	none	(2,3)	2
3	none	[]	none	none	2

We complete 2 assignments, so missed is 1. With k = 1 forgiveness, final loss is 0. The trace shows the greedy rule prioritizing tighter deadlines, which prevents losing the earliest-ending interval.

Example 2

Input:

n=4, d=2, k=0
(1,1), (1,1), (1,2), (2,2)

Day	Newly added	Heap after add	Removed expired	Chosen job	Done
1	all (start 1)	[1,1,2]	none	(1,1)	1
2	none	[1,2]	one expired 1	(2,2)	2

We complete 2 jobs. Two identical tight interval jobs compete for a single day, and only one can survive. This demonstrates the unavoidable loss when demand exceeds capacity.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Each job is inserted once into the heap and removed at most once, each operation logarithmic
Space	O(n)	Heap and job storage hold at most all active assignments

The constraints allow up to 2 × 10^5 total assignments, so an O(n log n) solution easily fits within both time and memory limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    import heapq

    t = int(input())
    out = []

    for _ in range(t):
        n, d, k = map(int, input().split())
        jobs = []
        for _ in range(n):
            s, e = map(int, input().split())
            jobs.append((s, e))

        jobs.sort()
        i = 0
        heap = []
        done = 0

        for day in range(1, d + 1):
            while i < n and jobs[i][0] <= day:
                heapq.heappush(heap, jobs[i][1])
                i += 1

            while heap and heap[0] < day:
                heapq.heappop(heap)

            if heap:
                heapq.heappop(heap)
                done += 1

        missed = n - done
        out.append(str(max(0, missed - k)))

    return "\n".join(out)

# provided sample-like checks
assert run("""1
1 1 0
1 1
""") == "0"

assert run("""1
3 2 1
1 1
1 2
2 2
""") == "0"

# tight overlap
assert run("""1
2 1 0
1 1
1 1
""") == "2"

# disjoint intervals
assert run("""1
3 3 0
1 1
2 2
3 3
""") == "0"

# wide intervals
assert run("""1
3 3 0
1 3
1 3
1 3
""") == "0"

Test input	Expected output	What it validates
1 1 0 / 1 1	0	single assignment base case
overlapping tight intervals	0	interaction with k forgiveness
two identical deadlines	2	over-capacity loss
disjoint intervals	0	optimal scheduling possible
all-wide intervals	0	greedy flexibility

Edge Cases

One edge case is when multiple assignments share the same single valid day. The heap will contain multiple identical end days, but only one can be popped per day. The algorithm naturally leaves the remaining ones in the heap until they expire, at which point they are discarded. For example:

n=2, d=1, k=0
(1,1), (1,1)

On day 1, both jobs are inserted. The heap contains [1,1]. One is chosen, leaving one uncompleted. On day 2 does not exist, so the remaining job is considered missed. Output is 2, matching the unavoidable constraint.

Another edge case occurs when intervals are very wide but start late. The algorithm correctly delays insertion until their start day, ensuring they are not prematurely considered. For instance:

n=1, d=5, k=0
(5,5)

No job is available until day 5, so it is only considered at the correct time. It is then scheduled immediately.