8 queens problem using backtracking

queens8.c

#include<stdio.h>
#include<stdlib.h>
int t[8] = {-1};
int sol = 1;

void printsol()
{
	int i,j;
	char crossboard[8][8];
	for(i=0;i<8;i++)
	{
		for(j=0;j<8;j++)
		{
			crossboard[i][j]='_';
		}
	}
	for(i=0;i<8;i++)
	{
			crossboard[i][t[i]]='q';
	}
	
	for(i=0;i<8;i++)
	{
		for(j=0;j<8;j++)
		{
			printf("%c ",crossboard[i][j]);
		}
		printf("\n");
	}  
}
int empty(int i)
{
	int j=0;
	while((t[i]!=t[j])&&(abs(t[i]-t[j])!=(i-j))&&j<8)j++;
	return i==j?1:0;
}

void queens(int i)
{
	for(t[i] = 0;t[i]<8;t[i]++)
	{
		if(empty(i))
		{
			if(i==7){
				printsol();
				printf("\n solution %d\n",sol++);
			}
			else
			queens(i+1);
		}
	}
}

	

int main()
{
	queens(0);
	return 0;
}

can you please explain me the code

I'm not the author but here is how I read this code:

• The array t holds in which position a queen stands in each row. For example t[0] = 0 means there's a queen in row 0 col 0.
• The key function here is empty() which checks if a queen can be placed at a certain position. t[i] != t[j] means they're in the same column, abs(t[i] - t[j]) != (i - j) means they align diagonally. Those are invalid positions.
• The queens() function tests all positions within a row. If the position is good, it calls itself for the next row. Notice that if the next row fails it will continue with the current row where it left off.

Nice job.