Problem J. Let Sudoku Rotate

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)

Total Submission(s):     Accepted Submission(s): 

Problem Description

Sudoku is a logic-based, combinatorial number-placement puzzle, which is popular around the world.

In this problem, let us focus on puzzles with × grids, which consist of × regions. The objective is to fill the whole grid with hexadecimal digits, i.e. 0123456789ABCDEF, so that each column, each row, and each region contains all hexadecimal digits. The figure below shows a solved sudoku.

Yesterday, Kazari solved a sudoku and left it on the desk. However, Minato played a joke with her - he performed the following operation several times.

* Choose a region and rotate it by  degrees counterclockwise.

She burst into tears as soon as she found the sudoku was broken because of rotations.

Could you let her know how many operations her brother performed at least?

Input

The first line of the input contains an integer T (≤T≤) denoting the number of test cases.

Each test case consists of exactly  lines with  characters each, describing a broken sudoku.

Output

For each test case, print a non-negative integer indicating the minimum possible number of operations.

Sample Input

681D5A0C9FDBB2F7

0A734B62E167D9E5

5C9B73EF3C208410

F24ED18948A5CA63

39FAED5616400B74

D120C4B7CA3DEF38

7EC829A085BE6D51

B56438F129F79C2A

5C7FBC4E3D08719F

AE8B1673BF42A58D

60D3AF25619C30BE

294190D8EA57264C

C7D1B35606835EAB

AF52A1E019BE4306

8B36DC78D425F7C9

E409492FC7FA18D2

Sample Output

Hint

The original sudoku is same as the example in the statement.

#include<iostream>

#include<stdio.h>

#include<cmath>

#include<algorithm>

#include<string.h>

using namespace std;

#define ll long long

int ans;

int a[][];

int tmp[][]={};

void rotat(int x,int y)

{

    for(int i=;i<=;i++)

        for(int j=;j<=;j++)

            tmp[j][-i+]=a[(x-)*+i][(y-)*+j];

    for(int i=;i<=;i++)

        for(int j=;j<=;j++)

            a[(x-)*+i][(y-)*+j]=tmp[i][j];

}

bool check(int x,int y)

{

    int book[];

    for(int i=x*-;i<=x*;i++)

    {

        memset(book,,sizeof book);

        for(int j=;j<=y*;j++)

        {

            if(!book[a[i][j]])

                book[a[i][j]]=;

            else

                return ;

        }

    }

    for(int j=y*-;j<=y*;j++)

    {

        memset(book,,sizeof book);

        for(int i=;i<=x*;i++)

        {

            if(!book[a[i][j]])

                book[a[i][j]]=;

            else

                return ;

        }

    }

    return ;

}

void bfs(int x,int y,int sum)

{

    if(x==)

    {

        if(ans>sum)

          ans=sum;

        return;

    }

    if(sum>ans)

    {

        return;

    }

    for(int i=;i<;i++)

    {

        if(i)  rotat(x,y);

        if(check(x,y))

        {

            if(y+<=)

            bfs(x,y+,sum+i);

            else

            bfs(x+,,sum+i);

        }

    }

    rotat(x,y);

}

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        ans=*;

        char str[];

        for(int i=;i<=;i++)

        {

            scanf("%s",str+);

            for(int j=;j<=;j++)

            {

                if(str[j]>'')

                a[i][j]=str[j]-'A'+;

                else a[i][j]=str[j]-'';

            }

        }

        bfs(,,);

        cout<<ans<<endl;

    }

    return ;

}