**Author:** Sergey Nagin

**Tester:** Istvan Nagy

**Editorialist:** Misha Chorniy

## Difficulty:

Medium-hard

# Pre-Requisites:

DP(Knapsack)

## Problem Statement

You are given a fragment of code and binary 3-dimensional array A[0..N-1, 0..N-1, 0..N-1], where N can be only power of 2, 1 <= N <= 32. Consider all 3-dimensional arrays, which can be created with changing no more than K elements of A and apply this code for them. After that, we are interested in finding the minimal and maximal value of resulting function over all such arrays. Below is the fragment of code given in the statement:

```
int F(box A, int dx = 0, int dy = 0, int dz = 0, int size = N) {
vector B = {};
for (int i = dx; i < dx + size; i++) {
for (int j = dy; j < dy + size; j++) {
for (int k = dz; k < dz + size; k++) {
B.push_back(A[i][j][k]);
}
}
}
sort(B.begin(), B.end());
if (B[0] == B[size * size * size - 1]) {
return 1;
}
int result = 0;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
result += f(A,
dx + i * size / 2,
dy + j * size / 2,
dz + k * size / 2,
size / 2);
}
}
}
return result;
}
```

## Explanation

## Subtask 1

For this subtask, we can observe that array A has N^3 binary cells. Let’s compress it in one-dimensional array B of size N^3. B_{i*N^2+j*N+k} = A_{i,j,k}, after that let’s iterate over all bitmasks. Assume that i-th element will be i-th bit in a mask, then we need to iterate in the range from 0 to 2^{N^3}-1, check the number of different bits if this number not more than K then launch the code above and relax the minimal and maximal values.

# Subtasks 2-5

Let’s try to understand what is the function F in the statement. Let’s make several observations:

What we can do with it? Let’s denote couple of functions:

In which cases maxValue[dx..dx+size-1,dy..dy+size-1,dz..dz+size-1][i] will be 1? Only in that case if i = 0 and all values inside cube are zeroes or ones. Because if i > 0, and all the values equal, we can change states exactly 1 of them, after that value of F will be recalculated using smaller sub-boxes.

In which cases minValue[dx..dx+size-1,dy..dy+size-1,dz..dz+size-1][i] will be 1? Only in those cases

when we can receive equal values inside the cube and change not more than i values inside of it. In other words, if

max(zeroes[dx..dx+size-1,dy..dy+size-1,dz..dz+size-1],

```
$ones[dx..dx+size-1,dy..dy+size-1,dz..dz+size-1])+i>=size^3$
```

Otherwise, how to recalculate value of minValue and maxValue using smaller sub-cubes and values of minValue and maxValue for them. It will be very similar to knapsack problem. minValue[cube][i] can be calculated if we don’t change more than i cells inside 8 smaller subcubes. Analogically with maxValue.

minValue[cube][i] = min(\sum_{i_{1}+i_{2}+i_{3}+i_{4}+i_{5}+i_{6}+i_{7}+i_{8}=i} minValue[subcube_{1}][i_{1}]+..+minValue[subcube_{8}][i_{8}])

maxValue[cube][i] = max(\sum_{i_{1}+i_{2}+i_{3}+i_{4}+i_{5}+i_{6}+i_{7}+i_{8}=i} maxValue[subcube_{1}][i_{1}]+..+maxValue[subcube_{8}][i_{8}])

It gives us the idea for the next solution:

```
go(dx, dy, dz, N) //function returns the pair of minValue and maxValue for cube
if N = 1 //If we have only one cell
return {1, 1}, //minValue[0], minValue[1]
{1, 1} //maxValue[0], maxValue[1]
minValue = array of size = N * N * N+1
maxValue = array of size = N * N * N+1
//N*N*N - is the number of cells which can be changed inside cube
was = 0
for i = 0..1 //Iterate over 8 subcubes
for j = 0..1
for k = 0..1
was += 1 //number of subcubes which was processed
tMinValue, tMaxValue = go(dx + i * (N / 2), dy + j * (N /2), dz + z * (N / 2), N / 2)
//Observe that tMinValue and tMaxValue has sizes N * N * N / 8 + 1
nMinValue = array(N * N * N / 8 * was + 1, +1000000)
nMaxValue = array(N * N * N / 8 * was + 1, -1000000)
//new values of minValue and maxValue, in optimized code, we can get rid of them
for v = 0..(N * N * N / 8) * was //combine values like a knapsack problem
for u = 0..min(N * N * N / 8, u) //iterate over values for subcube
nMinValue[v] = min(nMinValue[v], tMinValue[u] + minValue[v - u])
nMaxValue[v] = max(nMaxValue[v], tMaxValue[u] + maxValue[v - u])
for v = 0..(N * N * N / 8) * was
minValue[v] = nMinValue[v]
maxValue[v] = nMaxValue[v]
//Corner cases, not going into subcubes
if max(zeroes[dx..dx+N-1,dy..dy+N-1,dz..dz+N-1],ones[dx..dx+N-1,dy..dy+N-1,dz..dz+N-1]) == N * N * N
maxValue[0] = 1
for i = 0..N * N * N
if min(zeroes[dx..dx+N-1,dy..dy+N-1,dz..dz+N-1],ones[dx..dx+N-1,dy..dy+N-1,dz..dz+N-1]) + i >= N * N * N
minValue[i] = 1
return minValue, maxValue
```

Total complexity of this algorithm will be O(N*N*N * log(N) + N*N*N/8 * N*N*N/8 * log(N)) = O(N^6*log(N)), but with very small constant. There are many optimizations for it, the most crucial one is using a one-dimensional array instead three-dimensional.

## Solution:

Setter’s solution can be found here

Tester’s solution can be found here

**Please feel free to post comments if anything is not clear to you.**