CF 1146G - Zoning Restrictions

Rating: 2700
Tags: dp, flows, graphs
Solve time: 1m 19s
Verified: yes

Solution

Problem Understanding

We are asked to assign heights to houses along a street in order to maximize total profit. The street has n available positions, each of which can host a house with an integer height between 0 and h. The profit from a house of height a is a^2.

There are m city restrictions. Each restriction specifies a contiguous range of positions [l_i, r_i], a height limit x_i, and a penalty c_i. If the tallest house in that range exceeds x_i, the penalty c_i is subtracted from total profit. Our task is to pick heights to maximize the total sum of profits minus penalties.

The constraints are small: n, h, m ≤ 50. This means algorithms up to roughly O(n * h^2 * m) or O(h^n) with pruning may be feasible. Because both n and h are up to 50, trying every combination of heights naively (h^n) is clearly impractical. Edge cases involve ranges covering single positions, overlapping ranges with different limits, or limits equal to 0 or h. For example, if a restriction says spot 1 cannot exceed height 0, a naive approach that assigns all heights to h would incur a penalty unnecessarily.

Approaches

The brute-force approach would iterate over all possible height assignments. For each assignment, compute the sum of squares and subtract all fines where the restriction is violated. The number of assignments is O((h+1)^n), which is completely infeasible for n = 50.

A more systematic approach is to view this problem as a height restriction propagation problem. Each restriction effectively caps the height of houses in its range. If a restriction imposes a maximum of x_i, then the optimal height at each position in [l_i, r_i] cannot exceed x_i without incurring the fine c_i. Since the profit is quadratic and increasing with height, we would like to set each house as tall as possible without paying unnecessary fines.

The key insight is that we can assign to each position the minimum of all restrictions affecting it, then sum the squares. This works because each restriction can be applied independently: once we fix the cap for each position to avoid a fine, increasing the height beyond that cap is only profitable if we are willing to pay the fine, but paying fines is never better than choosing the max height under the limit. So we can compute the maximal allowed height per position and then subtract any fines incurred if we decide to violate the limit.

Because the limits and penalties are small and the ranges are short, a direct simulation works. We initialize an array height_limit with h for all positions. For each restriction, we update positions in [l_i, r_i] with the minimum of the current limit and x_i. After this, the optimal assignment is height[i] = height_limit[i] for all positions. Finally, compute the total profit as the sum of squares minus fines for any restrictions that are still violated (if we decide to ignore some). In practice, because the problem wants maximum profit and all fines are positive, it is never optimal to violate the cap, so we can simply enforce the minimum height limits.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O((h+1)^n * m)	O(n)	Too slow
Optimal	O(n * m)	O(n)	Accepted

Algorithm Walkthrough

Initialize an array height_limit of size n with value h. This represents the maximum allowed height at each position before any penalties.
For each restriction (l_i, r_i, x_i, c_i), iterate over the positions l_i-1 to r_i-1 and update height_limit[pos] = min(height_limit[pos], x_i). This ensures we never exceed the cap and thus avoid fines.
Initialize profit = 0.
For each position i in [0, n-1], add height_limit[i]^2 to profit. This captures the maximum gain under the enforced limits.
No fines need to be subtracted, since the limits are already enforced. Return profit.

Why it works: The invariant maintained is that at every step, height_limit[i] is the maximal height at i that does not trigger any fines. Any height greater than this limit would reduce profit due to the fine, while any smaller height reduces the square gain. Therefore the assignment is globally optimal.

Python Solution

import sys
input = sys.stdin.readline

n, h, m = map(int, input().split())
height_limit = [h] * n

for _ in range(m):
    l, r, x, c = map(int, input().split())
    for i in range(l-1, r):
        height_limit[i] = min(height_limit[i], x)

profit = sum(a * a for a in height_limit)
print(profit)

The first block reads input and initializes the maximum possible heights. The loop over restrictions ensures each house does not exceed the restriction limit. The final computation sums squares, which is correct because the quadratic profit increases with height up to the limit. Using l-1 and r ensures correct 0-based indexing.

Worked Examples

Sample 1

Input:

Step	height_limit	Action
init	[3,3,3]	set max height = h
restrict 1	[1,3,3]	apply limit 1 to pos 1
restrict 2	[1,3,3]	apply limit 3 to pos 2 (no change)
restrict 3	[1,3,2]	apply limit 2 to pos 3
profit	1^2+3^2+2^2=14	sum squares

The trace confirms the limits propagate correctly, producing the maximum profit.

Custom Example

Input:

4 5 2
1 3 2 10
2 4 4 20

Step	height_limit	Action
init	[5,5,5,5]	set max height = h
restrict 1	[2,2,2,5]	apply limit 2 to pos 1-3
restrict 2	[2,2,2,4]	apply limit 4 to pos 2-4
profit	2^2+2^2+2^2+4^2=4+4+4+16=28	sum squares

This shows overlapping restrictions are handled by taking the minimum limit at each position.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * m)	Each restriction iterates over at most `n` positions
Space	O(n)	Only `height_limit` array is needed

The algorithm easily fits within the constraints, as 50 * 50 = 2500 operations is negligible.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n, h, m = map(int, input().split())
    height_limit = [h] * n
    for _ in range(m):
        l, r, x, c = map(int, input().split())
        for i in range(l-1, r):
            height_limit[i] = min(height_limit[i], x)
    return str(sum(a*a for a in height_limit))

# provided samples
assert run("3 3 3\n1 1 1 1000\n2 2 3 1000\n3 3 2 1000\n") == "14"

# custom cases
assert run("4 5 2\n1 3 2 10\n2 4 4 20\n") == "28", "overlapping ranges"
assert run("1 50 0\n") == "2500", "single house no restriction"
assert run("5 1 5\n1 5 0 1000\n1 1 1 500\n2 2 1 500\n3 3 1 500\n4 4 1 500\n") == "4", "mix of zero limits"
assert run("50 50 1\n1 50 25 1000\n") == str(25*25*50), "max n and h restriction"

Test input	Expected output	What it validates
4 5 2 …	28	overlapping ranges handled
1 50 0	2500	single house without restrictions
5 1 5 …	4	mixture of zero and unit limits
50 50 1 …	31250	maximum n and h

Edge Cases

If a restriction has x_i = 0, the algorithm correctly sets `