RosettaCodeData/Task/N-queens-problem/C/n-queens-problem-3.c

92 lines
2.3 KiB
C

#include <stdio.h>
#include <stdlib.h>
typedef unsigned int uint;
uint count = 0;
#define ulen sizeof(uint) * 8
/* could have defined as int solve(...), but void may have less
chance to confuse poor optimizer */
void solve(int n)
{
int cnt = 0;
const uint full = -(int)(1 << (ulen - n));
register uint bits, pos, *m, d, e;
uint b0, b1, l[32], r[32], c[32], mm[33] = {0};
n -= 3;
/* require second queen to be left of the first queen, so
we ever only test half of the possible solutions. This
is why we can't handle n=1 here */
for (b0 = 1U << (ulen - n - 3); b0; b0 <<= 1) {
for (b1 = b0 << 2; b1; b1 <<= 1) {
d = n;
/* c: columns occupied by previous queens.
l: columns attacked by left diagonals
r: by right diagnoals */
c[n] = b0 | b1;
l[n] = (b0 << 2) | (b1 << 1);
r[n] = (b0 >> 2) | (b1 >> 1);
/* availabe columns on current row. m is stack */
bits = *(m = mm + 1) = full & ~(l[n] | r[n] | c[n]);
while (bits) {
/* d: depth, aka row. counting backwards
because !d is often faster than d != n */
while (d) {
/* pos is right most nonzero bit */
pos = -(int)bits & bits;
/* mark bit used. only put current bits
on stack if not zero, so backtracking
will skip exhausted rows (because reading
stack variable is sloooow compared to
registers) */
if ((bits &= ~pos))
*m++ = bits | d;
/* faster than l[d+1] = l[d]... */
e = d--;
l[d] = (l[e] | pos) << 1;
r[d] = (r[e] | pos) >> 1;
c[d] = c[e] | pos;
bits = full & ~(l[d] | r[d] | c[d]);
if (!bits) break;
if (!d) { cnt++; break; }
}
/* Bottom of stack m is a zero'd field acting
as sentinel. When saving to stack, left
27 bits are the available columns, while
right 5 bits is the depth. Hence solution
is limited to size 27 board -- not that it
matters in foreseeable future. */
d = (bits = *--m) & 31U;
bits &= ~31U;
}
}
}
count = cnt * 2;
}
int main(int c, char **v)
{
int nn;
if (c <= 1 || (nn = atoi(v[1])) <= 0) nn = 8;
if (nn > 27) {
fprintf(stderr, "Value too large, abort\n");
exit(1);
}
/* Can't solve size 1 board; might as well skip 2 and 3 */
if (nn < 4) count = nn == 1;
else solve(nn);
printf("\nSolutions: %d\n", count);
return 0;
}