POJ 3239 Solution to the n Queens Puzzle

/*回溯*/

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define N 20
int x[N]; //x[i]:i行的第x[i]列


int ok(int k)
{
    int i;
    for (i=1; i<k; i++)
    {
        if (x[i] == x[k] || abs(x[i]-x[k]) == abs(i-k))
            return 0;
    }
    return 1;
}

void solve(int n)
{
    int k;  //找第k行上的
    k = 1;
    x[k] = 0;
    while (k > 0)
    {
        x[k]++;
        while (x[k] <= n && !ok(k))
        {
            x[k]++;
        }
        if (x[k] <= n)
        {
         if (k == n)
                break;
            else
            {
                k++;
                x[k] = 0;
            }
        }
        else
        {
            x[k] = 0;
            k--;
        }
    }
}

int main()
{
    int n, i;
    while(scanf("%d", &n), n)
    {
        solve(n);
        for (i=1; i<=n; i++)
        {
            printf("%d ",x[i]);
        }
        printf("\n");
    }
    return 0;
}


n最大300,此方法肯定行不通， 只能用构造法，至于构造法...网上到处都是