Fishes
Contest: Codeforces Round 456 (Div. 2)
Difficulty rating: 2100
Tags: data structures, graphs, greedy, probabilities, shortest paths

While Grisha was celebrating New Year with Ded Moroz, Misha gifted Sasha a small rectangular pond of size n × m, divided into cells of size 1 × 1, inhabited by tiny evil fishes (no more than one fish per cell, otherwise they'll strife!).

The gift bundle also includes a square scoop of size r × r, designed for fishing. If the lower-left corner of the scoop-net is located at cell (x, y), all fishes inside the square (x, y)...(x + r - 1, y + r - 1) get caught. Note that the scoop-net should lie completely inside the pond when used.

Unfortunately, Sasha is not that skilled in fishing and hence throws the scoop randomly. In order to not frustrate Sasha, Misha decided to release k fishes into the empty pond in such a way that the expected value of the number of caught fishes is as high as possible. Help Misha! In other words, put k fishes in the pond into distinct cells in such a way that when the scoop-net is placed into a random position among (n - r + 1)·(m - r + 1) possible positions, the average number of caught fishes is as high as possible.

Input format:
The only line contains four integers n, m, r, k (1 ≤ n, m ≤ 105, 1 ≤ r ≤ min(n, m), 1 ≤ k ≤ min(n·m, 105)).

Output format:
Print a single number — the maximum possible expected number of caught fishes.

You answer is considered correct, is its absolute or relative error does not exceed 10 - 9. Namely, let your answer be a, and the jury's answer be b. Your answer is considered correct, if $$\frac{|a-b|}{\max(1, |b|)} \leq 10^{-9}$$.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first example you can put the fishes in cells (2, 1), (2, 2), (2, 3). In this case, for any of four possible positions of the scoop-net (highlighted with light green), the number of fishes inside is equal to two, and so is the expected value.

Example input:
3 3 2 3

Example output:
2.0000000000

Example input:
12 17 9 40

Example output:
32.8333333333

Editorial:
Let's solve a simplified problem first. Assume we know all fishes' positions (xi, yi) (1-indexed). Denote as g(x, y) the amount of fish that is inside a scoop with lower-left angle located at (x, y). Then the expected value is equal to:
It's quite obvious that straightforward computation will result into O(n·m) time. However, we can invert the problem and calculate for each fish f(xi, yi) — how many scoops it is covered by. f is given by the following formula:
Let's get back to $$\text{TE}$$ and transform it into:
In other words, in order to maximize the expected value we have to choose k best values among n·m possibilities.
From now on there are several approaches.
• Solution I.Imagine we're considering f(x0, y), i.e. with a fixed coordinate x0. Note that in this case f is y-convex, i.e. it's non-decreasing until some point, and after — non-increasing. Moreover, it always reaches its maximum at $$y = \left\lfloor \frac{m+1}{2} \right\rfloor$$. Denote the points to the left of this one D, and to the right (inclusive) — I.The rest of the solution looks as follows: initially we put 2·n points into the set, two per each x-coordinate, one for D and one for I. On each step we take the point with maximum value of f and replace it with its left or right neighbour (depending on which part it was from: D means that the substitute will be to the left, I — to the right).Complexity: $${\mathcal O}\left((n+k)\log n\right)$$.
• Solution II.Let's enhance the contemplations from the previous paragraph. Notice that f is convex by both values. Then we can take $$(\left\lfloor \frac{n+1}{2} \right\rfloor,\left\lfloor \frac{m+1}{2} \right\rfloor)$$) as the starting point and launch a breadth-first search with a priority queue, erasing the most significant point according to f from the queue and checking all of its yet unvisited neighbours in a 4-connected area.Complexity: $$O(k \log k)$$ with a greater constant compared with solution one.