Counting Special Sub-Cubes


Problem Statement :


Given an n*n*n cube, let f(x,y,z) (where 1 <= x,y,z <= n) denote the value stored in cell (x,y,z).

A k*k*k sub-cube (where 1 <= k <= n) of an n*n*n cube is considered to be special if the maximum value stored in any cell in the sub-cube is equal to k.

For each k in the inclusive range [1,n], calculate the number of special sub-cubes. Then print each (count)k as a single line of space-separated integers (i.e., count1,count2,...,countn).

Input Format

The first line contains an integer, q, denoting the number of queries. The 2.q subsequent lines describe each query over two lines:

1.The first line contains an integer, n, denoting the side length of the initial cube.
2.The second line contains n^3 space-separated integers describing an array of n^3 integers in the form a0,a1,...,a(n^3-1). The integer in some cell (x,y,z) is calculated using the formula a[(x-1).n^2 + (y-1).n + z].

Constraints
1 <= q <= 5
1 <= n <= 50
1 <= f(x,y,z) <=n where 1 <= x,y,z <=n
Output Format

For each query, print n space-separated integers where the ith integer denotes the number of special sub-cubes for k=i.



Solution :



title-img


                            Solution in C :

In C++ :





#include <iostream>
#include <iomanip>
#include <fstream>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>

using namespace std;
typedef long long ll;
const int MAXN = 52;

int Q;
int N;
int val[MAXN][MAXN][MAXN];
int dp[MAXN][MAXN][MAXN];

int main()
{
    cin >> Q;
    for (int test = 0; test < Q; test++)
    {

    cin >> N;
    for (int i = 0; i < MAXN; i++)
        for (int j = 0; j < MAXN; j++)
            for (int k = 0; k < MAXN; k++)
                val[i][j][k] = dp[i][j][k] = 0;

    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            for (int k = 0; k < N; k++)
            {
                cin >> val[i][j][k];
                dp[i][j][k] = val[i][j][k];
            }

    for (int i = 1; i <= N; i++)
    {
        int ans = 0;
        for (int j = 0; j <= N - i; j++)
            for (int k = 0; k <= N - i; k++)
                for (int l = 0; l <= N - i; l++)
                {
                    if (dp[j][k][l] == i) ans++;

                    for (int m = 0; m < 2; m++)
                        for (int n = 0; n < 2; n++)
                            for (int o = 0; o < 2; o++)
                                dp[j][k][l] = max (dp[j][k][l], dp[j+m][k+n][o+l]);
                
                }

        cout << ans;
        if (i < N)
            cout << " ";
    }
    cout << "\n";
    }
    return 0;
}








In Java :






import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 *
 * @author jin
 */
public class Solution {

    public static void main(String args[]) throws Exception {

        int q = input();
        for (int i = 0; i < q; i++) {
            int n = input();
            int ans[] = new int[n];

            int b = n * n * n, c = n * n;
            int arr[][][] = new int[50][50][50];
            int crr[][][] = new int[50][50][50];
            int ck[] = new int[b];
            input(ck, b);
            int count=0;
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    for (int l = 0; l < n; l++) {
                        arr[j][k][l] = ck[count++];
                //        System.out.print(arr[j][k][l]+" ");
                        if (arr[j][k][l] == 1) {
                            ans[0]++;
                        }

                    }
              //      System.out.println();
                }
            //    System.out.println("");

            }
            System.out.print(ans[0] + " ");
            for (int p = 1; p < n; p++) {

                for (int l = 0; l < n - p; l++) {
                    for (int j = 0; j < n; j++) {
                        for (int k = 0; k < n; k++) {
                            if (arr[j][k][l + 1] > arr[j][k][l]) {
                                arr[j][k][l] = arr[j][k][l + 1];
                            }

                        }

                    }
                }
                for (int l = 0; l < n - p; l++) {
                    for (int j = 0; j < n; j++) {
                        for (int k = 0; k < n-p; k++) {
                          //  crr[j][l][k] = arr[j][l][k];
                            if (arr[j][l + 1][k] > arr[j][l][k]) {
                                arr[j][l][k] = arr[j][l + 1][k];
                            }

                        }

                    }
                }
                for (int l = 0; l < n - p; l++) {
                    for (int j = 0; j < n-p; j++) {
                        for (int k = 0; k < n-p; k++) {
                            if (arr[l + 1][j][k] > arr[l][j][k]) {
                                arr[l][j][k] = arr[l + 1][j][k];
                            }

                        }

                    }
                }

                for (int l = 0; l < n - p; l++) {
                    for (int j = 0; j < n-p; j++) {
                        for (int k = 0; k < n-p; k++) {
                         //   System.out.println(l+" "+j+" "+k+"  "+(p+1));
                           if(arr[j][k][l]==p+1) 
                        ans[p]++;
                        }
                    }
                }

                System.out.print(ans[p] + " ");

            }
            System.out.println();
        }

    }

    static BufferedReader br = new BufferedReader(new InputStreamReader(
            System.in));
    private static String s[], w;

    public static void input(int a[], int p) throws IOException {
        s = br.readLine().split(" ");
        int i;
        for (i = 0; i < p; i++) {
            a[i] = Integer.parseInt(s[i]);
        }

    }

    public static void input(long a[], int p) throws IOException {
        s = br.readLine().split(" ");
        int i;
        for (i = 0; i < p; i++) {
            a[i] = Long.parseLong(s[i]);
        }
    }

    public static void input(double a[], int p) throws IOException {
        s = br.readLine().split(" ");
        int i;
        for (i = 0; i < p; i++) {
            a[i] = Double.parseDouble(s[i]);
        }
    }

    public static int input() throws IOException {
        int a;
        a = Integer.parseInt(br.readLine());
        return a;
    }

}








In C :





#include<stdio.h>


int cube(int x,int y,int z,int size,int count[],int table[50][50][50][50],int n,int arr[])
{
	if(table[x][y][z][size-1]==0)
	{
		if(size==1)
		   table[x][y][z][0] = arr[(x*n*n) + (y*n) +(z+1)];
		else
		{
			int max =  0;
			int a = cube(x,y,z,size-1,count,table,n,arr);
			if(a>max)
			max =a;
			int b = cube(x+1,y,z,size-1,count,table,n,arr);
			if(b>max)
			max  = b;
			int c = cube(x,y+1,z,size-1,count,table,n,arr);
			if(c>max)
			max =c;
			int d= cube(x+1,y+1,z,size-1,count,table,n,arr);
			if(d>max)
			max = d;
			int e= cube(x,y,z+1,size-1,count,table,n,arr);
			if(e>max)
			max = e;
			int f= cube(x+1,y,z+1,size-1,count,table,n,arr);
			if(f>max)
			max = f;
			int g= cube(x,y+1,z+1,size-1,count,table,n,arr);
			if(g>max)
			max = g;
			int h= cube(x+1,y+1,z+1,size-1,count,table,n,arr);
			if(h>max)
			max = h;
			table[x][y][z][size-1]=max;
		}
		if(table[x][y][z][size-1]==size)
		count[size-1]++;
	}
	return table[x][y][z][size-1];
}

int main()
{
  int query;
  scanf("%d",&query);
  while(query--)
  {
  	int n,i;
  scanf("%d",&n);
  int arr[(n*n*n)+1];
  for(i=1;i<=(n*n*n);i++)
  scanf("%d",&arr[i]);
  int table[50][50][50][50] = {0};
  int count[n];
  for(i=0;i<n;i++)
  count[i] =0;
  int p= cube(0,0,0,n,count,table,n,arr);
  for(i=0;i<n;i++)
  printf("%d ",count[i]);
  printf("\n");
  }
}








In Python3 :





def findMax(arr, s, indices, n):
    largest = 1
    for i in indices:
        cur = arr[s+i]
        if cur > largest:
            largest = cur
        elif cur == n:
            largest = n
            break
    return largest

def findSpecial(arr, k, n):
    temp = arr
    if k > 1:
        temp = []
        m = n - k + 2
        m2 = m*m
        indices = [0, 1, m, m+1, m2, m2+1, m2+m, m2+m+1]
        for i in range(m-1):
            start = m2*i
            for j in range(m-1):
                for z in range(m-1):
                    largest = findMax(arr, start, indices, n)
                    temp.append(largest)
                    start += 1
                start += 1
    return temp    
               
q = int(input())
for _ in range(q):
    n = int(input())
    arr = list(map(int, input().split()))
    for k in range(1, n+1):
        arr = findSpecial(arr, k, n)
        print(arr.count(k), end=' ')
    print('')
                        








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