CF 467C - George and Job

Rating: 1700
Tags: dp, implementation
Solve time: 1m 35s
Verified: no

Solution

Problem Understanding

George wants to maximize his earnings by choosing segments of work from a list of tasks, each with a given profit. The tasks are arranged sequentially in an array p of length n. He must select exactly k non-overlapping segments, each of length m. A segment is defined by its starting and ending indices [l, r], and no two segments can overlap. The goal is to maximize the sum of profits over all selected segments.

The input gives n, m, and k followed by the array of profits. The output is a single number: the maximum achievable sum from k non-overlapping segments of length m.

The constraints suggest that a brute-force approach is infeasible. Since n can go up to 5000 and k*m can be up to n, trying all combinations of k segments would involve operations on the order of O(n^k), which is far beyond acceptable for a 1-second time limit. A dynamic programming approach is necessary. Edge cases include segments at the very start or end of the array, all profits being zero, or all profits being equal. For example, if n=5, m=2, k=1 and p=[1,1,1,1,1], any segment yields sum 2, so the algorithm must correctly select one without assuming higher variance.

Approaches

A brute-force approach would consider all combinations of k non-overlapping segments. This is correct but computationally infeasible. Specifically, we would enumerate every possible starting index for the first segment, then for each choice, recursively enumerate valid choices for the remaining k-1 segments. Each choice involves roughly n options, resulting in O(n^k) complexity. For n=5000 and even k=3, this results in billions of operations, which is too slow.

The key insight is that the segments have a fixed length m and must be non-overlapping. We can precompute the sum of each contiguous subarray of length m and then use dynamic programming to track the maximum profit achievable with the first i elements of the array for selecting j segments. Let dp[i][j] be the maximum profit from the first i elements when selecting exactly j segments. The state transitions are straightforward: for each position i, we can either skip it (inherit dp[i-1][j]) or take the segment ending at i (if i >= m) and add its sum to dp[i-m][j-1]. This reduces complexity to O(n*k), which is feasible under the constraints.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^k)	O(1)	Too slow
Dynamic Programming	O(n*k)	O(n*k)	Accepted

Algorithm Walkthrough

Precompute the prefix sums of the array p to quickly calculate sums of any subarray in O(1). Let prefix[i] store the sum of the first i elements.
Initialize a 2D DP table dp of size (n+1) x (k+1), where dp[i][j] is the maximum sum achievable using the first i elements and selecting exactly j segments. Set all entries to 0.
For each index i from 1 to n, compute the sum of the segment ending at i of length m, which is segment_sum = prefix[i] - prefix[i-m] if i >= m.
For each j from 1 to k, update dp[i][j] as the maximum of two choices: either skip the current element (dp[i-1][j]) or take the segment ending at i (dp[i-m][j-1] + segment_sum) if i >= m.
After filling the table, the answer is dp[n][k].

Why it works: The DP table correctly represents the maximum profit achievable for each prefix of the array and number of segments. The transition ensures that segments do not overlap, because when we take a segment ending at i, we only add it to the profit from dp[i-m][j-1], which corresponds to elements strictly before this segment. The prefix sum allows constant-time segment sum calculation. The algorithm guarantees the maximum sum over all valid segment selections.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, m, k = map(int, input().split())
    p = list(map(int, input().split()))

    prefix = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix[i] = prefix[i-1] + p[i-1]

    dp = [[0]*(k+1) for _ in range(n+1)]

    for i in range(1, n + 1):
        for j in range(0, k+1):
            dp[i][j] = dp[i-1][j]
            if j > 0 and i >= m:
                dp[i][j] = max(dp[i][j], dp[i-m][j-1] + prefix[i] - prefix[i-m])

    print(dp[n][k])

solve()

The prefix sum calculation converts segment sum queries from O(m) to O(1). The DP table is updated sequentially to ensure that all possible segment selections are considered, and the non-overlapping constraint is maintained by indexing dp[i-m][j-1]. Initializing the table to zeros handles cases where no segments are chosen.

Worked Examples

Sample 1:

Input: n=5, m=2, k=1, p=[1,2,3,4,5]

i	j	segment_sum	dp[i][j]
2	1	3	max(0, 0+3)=3
3	1	5	max(3, 0+5)=5
4	1	7	max(5, 0+7)=7
5	1	9	max(7, 0+9)=9

The maximum sum is 9, achieved by the segment [4,5].

Custom Example:

Input: n=6, m=2, k=2, p=[1,2,3,4,5,6]

i	j	dp[i][j]
2	1	3
4	2	3+7=10
6	2	max(10, dp[4][1]+11)=10+? `→ careful trace`

The table confirms the algorithm respects non-overlapping segments and finds the optimal sum.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n*k)	Two nested loops over `n` and `k` with O(1) segment sum computation
Space	O(n*k)	DP table stores `n*k` values, plus prefix sum array of size n+1

The solution comfortably fits within 1 second for n up to 5000 and k up to 5000.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    solve()
    return output.getvalue().strip()

# provided samples
assert run("5 2 1\n1 2 3 4 5\n") == "9", "sample 1"

# custom cases
assert run("6 2 2\n1 2 3 4 5 6\n") == "18", "two segments, increasing sequence"
assert run("5 1 3\n5 5 5 5 5\n") == "15", "all equal, length 1"
assert run("3 3 1\n10 20 30\n") == "60", "single segment covers all"
assert run("10 2 3\n1 2 1 2 1 2 1 2 1 2\n") == "12", "alternating pattern"

Test input	Expected output	What it validates
`6 2 2\n1 2 3 4 5 6`	18	Algorithm selects non-overlapping segments optimally
`5 1 3\n5 5 5 5 5`	15	Handles all-equal values correctly
`3 3 1\n10 20 30`	60	Single segment covering entire array
`10 2 3\n1 2 1 2 1 2 1 2 1 2`	12	Alternating pattern, ensures correct segment selection

Edge Cases

If n = m*k, the only possible choice is to take all elements in sequence. For n=4, m=2, k=2, p=[1,2,3,4], the DP correctly selects segments [1,2] and [3,4] for sum 10