当为n-
Queen问题的所有可能解决方案实现算法时，我发现许多分支机构都可以达到相同的解决方案。有什么好的方法可以生成解决n皇后问题的每一个独特的解决方案？如何避免由不同分支机构（存储和比较除外）生成的重复解决方案？

这是我尝试的第一个解决方案：http :
//www.ideone.com/hDpr3

码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

/* crude */

#define QUEEN 'Q'
#define BLANK '.'

int is_valid (char **board, int n, int a, int b)
{
  int i, j;

  for (i=0; i<n; i++)
  {
    if (board[a][i] == QUEEN)
      return 0;
    if (board[i][b] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i>=0) && (j>=0); i--, j--)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i<n) && (j<n); i++, j++)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i>=0) && (j<n); i--, j++)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }

  for (i=a, j=b; (i<n) && (j>=0); i++, j--)
  {
    if (board[i][j] == QUEEN)
      return 0;
  }
  return 1;
}

void show_board (char **board, int n)
{
  int i, j;

  for (i=0; i<n; i++)
  {
    printf ("\n");
    for (j=0; j<n; j++)
    {
      printf (" %c", board[i][j]);
    }
  }
}

int nqdfs_all (char **board, int n, int d)
{
  int i, j, ret = 0;

  /* the last queen was placed on the last depth
   * therefore this dfs branch in the recursion 
   * tree is a solution, return 1
   */
  if (d == n)
  {
    /* Print whenever we find one solution */
    printf ("\n");
    show_board (board, n);
    return 1;
  }

  /* check all position */
  for (i=0; i<n; i++)
  {
    for (j=0; j<n; j++)
    {
      if (is_valid (board, n, i, j))
      {
    board[i][j] = QUEEN;
    nqdfs_all (board, n, d + 1);
    board[i][j] = BLANK;
      }
    }
  }

  return ret;  
}

int nqdfs_first (char **board, int n, int d)
{
  int i, j, ret = 0;

  /* the last queen was placed on the last depth
   * therefore this dfs branch in the recursion 
   * tree is a solution, return 1
   */
  if (d == n)
    return 1;

  /* check all position */
  for (i=0; i<n; i++)
  {
    for (j=0; j<n; j++)
    {
      if (is_valid (board, n, i, j))
      {
    board[i][j] = QUEEN;
    ret = nqdfs_first (board, n, d + 1);
    if (ret)
    {
      /* if the first branch is found, tell about its 
       * success to its parent, we will not look in other
       * solutions in this function.
       */
      return ret;
    }
    else
    {
      /* this was not a successful path, so replace the
       * queen with a blank, and prepare board for next
       * pass
       */
      board[i][j] = BLANK;
    }
      }
    }
  }

  return ret;
}

int main (void)
{
  char **board;
  int n, i, j, ret;

  printf ("\nEnter \"n\": ");
  scanf ("%d", &n);

  board = malloc (sizeof (char *) * n);
  for (i=0; i<n; i++)
  {
    board[i] = malloc (sizeof (char) * n);
    memset (board[i], BLANK, n * sizeof (char));
  }

  nqdfs_first (board, n, 0);
  show_board (board, n);

  printf ("\n");
  return 0;
}

为了生成所有可能的解决方案，我使用了相同的代码nqdfs_all ()功能，但没有将控件返回给父级，而是继续枚举和检查。对该函数的调用将显示不同分支获得的重复结果。