拉丁方阵

#include <iostream>
using namespace std;

const int N = 4;
int count = 0;
int result_num = 0;
int a[N][N];

void latin(int, int);
void print();
void clear();
bool ok();

void main()
{
    clear();
    latin(0,0);
    cout << "There are " << result_num << " results" << endl;
}

void latin(int x, int y)
{
    if (count == N * N)
    {
        print();
        ++result_num;
    }
    else
    {
        for (int i = 1; i <= N; ++i)
        {
            a[x][y] = i;
            ++count;
            if (ok())
            {
                int yy = (y + 1) % N;
                int xx = x;
                if (y == N - 1)
                    xx = x + 1;
               
                latin(xx, yy);
            }
            --count;
        }
        a[x][y] = 0;
    }
   
}

void print()
{
    for (int i = 0; i != N; ++i)
    {
        for (int j = 0; j != N; ++ j)
        {
            cout << a[i][j] << ' ';
        }
        cout << endl;
    }
    cout << endl;
}

void clear()
{
    for (int i = 0; i != N; ++i)
    {
        for (int j = 0; j != N; ++ j)
        {
            a[i][j] = 0;
        }
    }
    count = 0;
}

bool ok()
{
    for (int i = 0; i != N; ++i)
    {
        for (int j = 0; j != N; ++j)
        {
            if (a[i][j] == 0)
                continue;
            int row = i,col = j;
            for (int k = 0; k != N; ++k)
            {
                if (k != j)
                {
                    if (a[row][k] == a[i][j])
                        return false;
                }
            }
            for (k = 0; k != N; ++k)
            {
                if (k != i)
                {
                    if (a[k][col] == a[i][j])
                        return false;
                }
            }
        }
    }
    return true;
}