回溯算法:N皇后问题解析和多语言代码实现

2021年3月11日17:14:37 发表评论 748 次浏览

本文概述

我们已经讨论了奈特的《迷宫之旅》中的《骑士之旅》和《老鼠》套装1和套装2分别。让我们讨论N Queen作为可以使用回溯解决的另一个示例问题。

N Queen是在N×N棋盘上放置N个国际象棋皇后的问题, 这样就不会有两个女王互相攻击。例如, 以下是4 Queen问题的解决方案。

N皇后问题回溯31

预期的输出是一个二进制矩阵, 其中放置了皇后的块的二进制数为1。例如, 以下是上述4个女王解决方案的输出矩阵。

{ 0, 1, 0, 0}
              { 0, 0, 0, 1}
              { 1, 0, 0, 0}
              { 0, 0, 1, 0}

推荐:请在"实践首先, 在继续解决方案之前。

天真的算法

在船上生成皇后区的所有可能配置, 并打印满足给定约束的配置。

while there are untried configurations
{
   generate the next configuration
   if queens don't attack in this configuration then
   {
      print this configuration;
   }
}

回溯算法

这个想法是将皇后区从最左边的列开始一个一列地放置在不同的列中。当我们在一个列中放置一个皇后时, 我们检查是否与已经放置的皇后发生冲突。在当前列中, 如果找到没有冲突的行, 则将该行和列标记为解决方案的一部分。如果由于冲突而找不到这样的行, 那么我们将回溯并返回false。

1) Start in the leftmost column
2) If all queens are placed
    return true
3) Try all rows in the current column. 
   Do following for every tried row.
    a) If the queen can be placed safely in this row 
       then mark this [row, column] as part of the 
       solution and recursively check if placing
       queen here leads to a solution.
    b) If placing the queen in [row, column] leads to
       a solution then return true.
    c) If placing queen doesn't lead to a solution then
       unmark this [row, column] (Backtrack) and go to 
       step (a) to try other rows.
3) If all rows have been tried and nothing worked, return false to trigger backtracking.

回溯解决方案的实施

C / C ++

/* C/C++ program to solve N Queen Problem using
    backtracking */
#define N 4
#include <stdbool.h>
#include <stdio.h>
  
/* A utility function to print solution */
void printSolution( int board[N][N])
{
     for ( int i = 0; i < N; i++) {
         for ( int j = 0; j < N; j++)
             printf ( " %d " , board[i][j]);
         printf ( "\n" );
     }
}
  
/* A utility function to check if a queen can
    be placed on board[row][col]. Note that this
    function is called when "col" queens are
    already placed in columns from 0 to col -1.
    So we need to check only left side for
    attacking queens */
bool isSafe( int board[N][N], int row, int col)
{
     int i, j;
  
     /* Check this row on left side */
     for (i = 0; i < col; i++)
         if (board[row][i])
             return false ;
  
     /* Check upper diagonal on left side */
     for (i = row, j = col; i >= 0 && j >= 0; i--, j--)
         if (board[i][j])
             return false ;
  
     /* Check lower diagonal on left side */
     for (i = row, j = col; j >= 0 && i < N; i++, j--)
         if (board[i][j])
             return false ;
  
     return true ;
}
  
/* A recursive utility function to solve N
    Queen problem */
bool solveNQUtil( int board[N][N], int col)
{
     /* base case: If all queens are placed
       then return true */
     if (col >= N)
         return true ;
  
     /* Consider this column and try placing
        this queen in all rows one by one */
     for ( int i = 0; i < N; i++) {
         /* Check if the queen can be placed on
           board[i][col] */
         if (isSafe(board, i, col)) {
             /* Place this queen in board[i][col] */
             board[i][col] = 1;
  
             /* recur to place rest of the queens */
             if (solveNQUtil(board, col + 1))
                 return true ;
  
             /* If placing queen in board[i][col]
                doesn't lead to a solution, then
                remove queen from board[i][col] */
             board[i][col] = 0; // BACKTRACK
         }
     }
  
     /* If the queen cannot be placed in any row in
         this colum col  then return false */
     return false ;
}
  
/* This function solves the N Queen problem using
    Backtracking. It mainly uses solveNQUtil() to 
    solve the problem. It returns false if queens
    cannot be placed, otherwise, return true and
    prints placement of queens in the form of 1s.
    Please note that there may be more than one
    solutions, this function prints one  of the
    feasible solutions.*/
bool solveNQ()
{
     int board[N][N] = { { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 } };
  
     if (solveNQUtil(board, 0) == false ) {
         printf ( "Solution does not exist" );
         return false ;
     }
  
     printSolution(board);
     return true ;
}
  
// driver program to test above function
int main()
{
     solveNQ();
     return 0;
}

Java

/* Java program to solve N Queen Problem using
    backtracking */
public class NQueenProblem {
     final int N = 4 ;
  
     /* A utility function to print solution */
     void printSolution( int board[][])
     {
         for ( int i = 0 ; i < N; i++) {
             for ( int j = 0 ; j < N; j++)
                 System.out.print( " " + board[i][j]
                                  + " " );
             System.out.println();
         }
     }
  
     /* A utility function to check if a queen can
        be placed on board[row][col]. Note that this
        function is called when "col" queens are already
        placeed in columns from 0 to col -1. So we need
        to check only left side for attacking queens */
     boolean isSafe( int board[][], int row, int col)
     {
         int i, j;
  
         /* Check this row on left side */
         for (i = 0 ; i < col; i++)
             if (board[row][i] == 1 )
                 return false ;
  
         /* Check upper diagonal on left side */
         for (i = row, j = col; i >= 0 && j >= 0 ; i--, j--)
             if (board[i][j] == 1 )
                 return false ;
  
         /* Check lower diagonal on left side */
         for (i = row, j = col; j >= 0 && i < N; i++, j--)
             if (board[i][j] == 1 )
                 return false ;
  
         return true ;
     }
  
     /* A recursive utility function to solve N
        Queen problem */
     boolean solveNQUtil( int board[][], int col)
     {
         /* base case: If all queens are placed
            then return true */
         if (col >= N)
             return true ;
  
         /* Consider this column and try placing
            this queen in all rows one by one */
         for ( int i = 0 ; i < N; i++) {
             /* Check if the queen can be placed on
                board[i][col] */
             if (isSafe(board, i, col)) {
                 /* Place this queen in board[i][col] */
                 board[i][col] = 1 ;
  
                 /* recur to place rest of the queens */
                 if (solveNQUtil(board, col + 1 ) == true )
                     return true ;
  
                 /* If placing queen in board[i][col]
                    doesn't lead to a solution then
                    remove queen from board[i][col] */
                 board[i][col] = 0 ; // BACKTRACK
             }
         }
  
         /* If the queen can not be placed in any row in
            this colum col, then return false */
         return false ;
     }
  
     /* This function solves the N Queen problem using
        Backtracking.  It mainly uses solveNQUtil () to
        solve the problem. It returns false if queens
        cannot be placed, otherwise, return true and
        prints placement of queens in the form of 1s.
        Please note that there may be more than one
        solutions, this function prints one of the
        feasible solutions.*/
     boolean solveNQ()
     {
         int board[][] = { { 0 , 0 , 0 , 0 }, { 0 , 0 , 0 , 0 }, { 0 , 0 , 0 , 0 }, { 0 , 0 , 0 , 0 } };
  
         if (solveNQUtil(board, 0 ) == false ) {
             System.out.print( "Solution does not exist" );
             return false ;
         }
  
         printSolution(board);
         return true ;
     }
  
     // driver program to test above function
     public static void main(String args[])
     {
         NQueenProblem Queen = new NQueenProblem();
         Queen.solveNQ();
     }
}
// This code is contributed by Abhishek Shankhadhar

Python3

# Python3 program to solve N Queen 
# Problem using backtracking
global N
N = 4
  
def printSolution(board):
     for i in range (N):
         for j in range (N):
             print (board[i][j], end = " " )
         print ()
  
# A utility function to check if a queen can
# be placed on board[row][col]. Note that this
# function is called when "col" queens are
# already placed in columns from 0 to col -1.
# So we need to check only left side for
# attacking queens
def isSafe(board, row, col):
  
     # Check this row on left side
     for i in range (col):
         if board[row][i] = = 1 :
             return False
  
     # Check upper diagonal on left side
     for i, j in zip ( range (row, - 1 , - 1 ), range (col, - 1 , - 1 )):
         if board[i][j] = = 1 :
             return False
  
     # Check lower diagonal on left side
     for i, j in zip ( range (row, N, 1 ), range (col, - 1 , - 1 )):
         if board[i][j] = = 1 :
             return False
  
     return True
  
def solveNQUtil(board, col):
      
     # base case: If all queens are placed
     # then return true
     if col > = N:
         return True
  
     # Consider this column and try placing
     # this queen in all rows one by one
     for i in range (N):
  
         if isSafe(board, i, col):
              
             # Place this queen in board[i][col]
             board[i][col] = 1
  
             # recur to place rest of the queens
             if solveNQUtil(board, col + 1 ) = = True :
                 return True
  
             # If placing queen in board[i][col
             # doesn't lead to a solution, then
             # queen from board[i][col]
             board[i][col] = 0
  
     # if the queen can not be placed in any row in
     # this colum col then return false
     return False
  
# This function solves the N Queen problem using
# Backtracking. It mainly uses solveNQUtil() to
# solve the problem. It returns false if queens
# cannot be placed, otherwise return true and
# placement of queens in the form of 1s.
# note that there may be more than one
# solutions, this function prints one of the
# feasible solutions.
def solveNQ():
     board = [ [ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ] ]
  
     if solveNQUtil(board, 0 ) = = False :
         print ( "Solution does not exist" )
         return False
  
     printSolution(board)
     return True
  
# Driver Code
solveNQ()
  
# This code is contributed by Divyanshu Mehta

C#

// C# program to solve N Queen Problem 
// using backtracking 
using System;
      
class GFG 
{
     readonly int N = 4;
  
     /* A utility function to print solution */
     void printSolution( int [, ]board)
     {
         for ( int i = 0; i < N; i++) 
         {
             for ( int j = 0; j < N; j++)
                 Console.Write( " " + board[i, j]
                                   + " " );
             Console.WriteLine();
         }
     }
  
     /* A utility function to check if a queen can
     be placed on board[row, col]. Note that this
     function is called when "col" queens are already
     placeed in columns from 0 to col -1. So we need
     to check only left side for attacking queens */
     bool isSafe( int [, ]board, int row, int col)
     {
         int i, j;
  
         /* Check this row on left side */
         for (i = 0; i < col; i++)
             if (board[row, i] == 1)
                 return false ;
  
         /* Check upper diagonal on left side */
         for (i = row, j = col; i >= 0 && 
              j >= 0; i--, j--)
             if (board[i, j] == 1)
                 return false ;
  
         /* Check lower diagonal on left side */
         for (i = row, j = col; j >= 0 && 
                       i < N; i++, j--)
             if (board[i, j] == 1)
                 return false ;
  
         return true ;
     }
  
     /* A recursive utility function to solve N
     Queen problem */
     bool solveNQUtil( int [, ]board, int col)
     {
         /* base case: If all queens are placed
         then return true */
         if (col >= N)
             return true ;
  
         /* Consider this column and try placing
         this queen in all rows one by one */
         for ( int i = 0; i < N; i++) 
         {
             /* Check if the queen can be placed on
             board[i, col] */
             if (isSafe(board, i, col))
             {
                 /* Place this queen in board[i, col] */
                 board[i, col] = 1;
  
                 /* recur to place rest of the queens */
                 if (solveNQUtil(board, col + 1) == true )
                     return true ;
  
                 /* If placing queen in board[i, col]
                 doesn't lead to a solution then
                 remove queen from board[i, col] */
                 board[i, col] = 0; // BACKTRACK
             }
         }
  
         /* If the queen can not be placed in any row in
         this colum col, then return false */
         return false ;
     }
  
     /* This function solves the N Queen problem using
     Backtracking. It mainly uses solveNQUtil () to
     solve the problem. It returns false if queens
     cannot be placed, otherwise, return true and
     prints placement of queens in the form of 1s.
     Please note that there may be more than one
     solutions, this function prints one of the
     feasible solutions.*/
     bool solveNQ()
     {
         int [, ]board = {{ 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }};
  
         if (solveNQUtil(board, 0) == false )
         {
             Console.Write( "Solution does not exist" );
             return false ;
         }
  
         printSolution(board);
         return true ;
     }
  
     // Driver Code
     public static void Main(String []args)
     {
         GFG Queen = new GFG();
         Queen.solveNQ();
     }
}
  
// This code is contributed by Princi Singh

输出如下:

1值表示皇后区的位置

0  0  1  0 
 1  0  0  0 
 0  0  0  1 
 0  1  0  0

is_safe()函数中的优化

这个想法不是检查左右对角线的每个元素, 而是使用对角线的属性:

1. i和j的和是恒定的, 并且对于每个右对角线都是唯一的, 其中i是元素的行, j是元素的行

元素列。

2. i和j的差是恒定的, 并且对于每个左对角线都是唯一的, 其中i和j分别是元素的行和列。

回溯解决方案的实现(优化)

C / C ++

/* C/C++ program to solve N Queen Problem using
    backtracking */
#define N 4
#include <stdbool.h>
#include <stdio.h>
/* ld is an array where its indices indicate row-col+N-1
  (N-1) is for shifting the difference to store negative 
  indices */
int ld[30] = { 0 };
/* rd is an array where its indices indicate row+col
    and used to check whether a queen can be placed on 
    right diagonal or not*/
int rd[30] = { 0 };
/*column array where its indices indicates column and 
   used to check whether a queen can be placed in that
     row or not*/
int cl[30] = { 0 };
/* A utility function to print solution */
void printSolution( int board[N][N])
{
     for ( int i = 0; i < N; i++) {
         for ( int j = 0; j < N; j++)
             printf ( " %d " , board[i][j]);
         printf ( "\n" );
     }
}
  
/* A recursive utility function to solve N
    Queen problem */
bool solveNQUtil( int board[N][N], int col)
{
     /* base case: If all queens are placed
       then return true */
     if (col >= N)
         return true ;
  
     /* Consider this column and try placing
        this queen in all rows one by one */
     for ( int i = 0; i < N; i++) {
         /* Check if the queen can be placed on
           board[i][col] */
         /* A check if a queen can be placed on 
            board[row][col].We just need to check
            ld[row-col+n-1] and rd[row+coln] where
            ld and rd are for left and right 
            diagonal respectively*/
         if ((ld[i - col + N - 1] != 1 &&
                   rd[i + col] != 1) && cl[i] != 1) {
             /* Place this queen in board[i][col] */
             board[i][col] = 1;
             ld[i - col + N - 1] =
                           rd[i + col] = cl[i] = 1;
  
             /* recur to place rest of the queens */
             if (solveNQUtil(board, col + 1))
                 return true ;
  
             /* If placing queen in board[i][col]
                doesn't lead to a solution, then
                remove queen from board[i][col] */
             board[i][col] = 0; // BACKTRACK
             ld[i - col + N - 1] =
                          rd[i + col] = cl[i] = 0;
         }
     }
  
     /* If the queen cannot be placed in any row in
         this colum col  then return false */
     return false ;
}
/* This function solves the N Queen problem using
    Backtracking. It mainly uses solveNQUtil() to 
    solve the problem. It returns false if queens
    cannot be placed, otherwise, return true and
    prints placement of queens in the form of 1s.
    Please note that there may be more than one
    solutions, this function prints one  of the
    feasible solutions.*/
bool solveNQ()
{
     int board[N][N] = { { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 } };
  
     if (solveNQUtil(board, 0) == false ) {
         printf ( "Solution does not exist" );
         return false ;
     }
  
     printSolution(board);
     return true ;
}
  
// driver program to test above function
int main()
{
     solveNQ();
     return 0;
}

Java

/* Java program to solve N Queen Problem 
using backtracking */
import java.util.*;
  
class GFG 
{
static int N = 4 ;
  
/* ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative 
indices */
static int []ld = new int [ 30 ];
  
/* rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on 
right diagonal or not*/
static int []rd = new int [ 30 ];
  
/*column array where its indices indicates column and 
used to check whether a queen can be placed in that
     row or not*/
static int []cl = new int [ 30 ];
  
/* A utility function to print solution */
static void printSolution( int board[][])
{
     for ( int i = 0 ; i < N; i++)
     {
         for ( int j = 0 ; j < N; j++)
             System.out.printf( " %d " , board[i][j]);
         System.out.printf( "\n" );
     }
}
  
/* A recursive utility function to solve N
Queen problem */
static boolean solveNQUtil( int board[][], int col)
{
     /* base case: If all queens are placed
     then return true */
     if (col >= N)
         return true ;
  
     /* Consider this column and try placing
     this queen in all rows one by one */
     for ( int i = 0 ; i < N; i++)
     {
          
         /* Check if the queen can be placed on
         board[i][col] */
         /* A check if a queen can be placed on 
         board[row][col].We just need to check
         ld[row-col+n-1] and rd[row+coln] where
         ld and rd are for left and right 
         diagonal respectively*/
         if ((ld[i - col + N - 1 ] != 1 &&
              rd[i + col] != 1 ) && cl[i] != 1 )
         {
             /* Place this queen in board[i][col] */
             board[i][col] = 1 ;
             ld[i - col + N - 1 ] =
             rd[i + col] = cl[i] = 1 ;
  
             /* recur to place rest of the queens */
             if (solveNQUtil(board, col + 1 ))
                 return true ;
  
             /* If placing queen in board[i][col]
             doesn't lead to a solution, then
             remove queen from board[i][col] */
             board[i][col] = 0 ; // BACKTRACK
             ld[i - col + N - 1 ] =
             rd[i + col] = cl[i] = 0 ;
         }
     }
  
     /* If the queen cannot be placed in any row in
         this colum col then return false */
     return false ;
}
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to 
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
static boolean solveNQ()
{
     int board[][] = {{ 0 , 0 , 0 , 0 }, { 0 , 0 , 0 , 0 }, { 0 , 0 , 0 , 0 }, { 0 , 0 , 0 , 0 }};
  
     if (solveNQUtil(board, 0 ) == false ) 
     {
         System.out.printf( "Solution does not exist" );
         return false ;
     }
  
     printSolution(board);
     return true ;
}
  
// Driver Code
public static void main(String[] args)
{
     solveNQ();
}
}
  
// This code is contributed by Princi Singh

Python3

""" Python3 program to solve N Queen Problem using 
backtracking """
N = 4
  
""" ld is an array where its indices indicate row-col+N-1 
(N-1) is for shifting the difference to store negative 
indices """
ld = [ 0 ] * 30
  
""" rd is an array where its indices indicate row+col 
and used to check whether a queen can be placed on 
right diagonal or not"""
rd = [ 0 ] * 30
  
"""column array where its indices indicates column and 
used to check whether a queen can be placed in that 
     row or not"""
cl = [ 0 ] * 30
  
""" A utility function to print solution """
def printSolution(board): 
     for i in range (N):
         for j in range (N):
             print (board[i][j], end = " " )
         print () 
  
""" A recursive utility function to solve N 
Queen problem """
def solveNQUtil(board, col): 
      
     """ base case: If all queens are placed
         then return True """
     if (col > = N):
         return True
          
     """ Consider this column and try placing
         this queen in all rows one by one """
     for i in range (N):
          
         """ Check if the queen can be placed on board[i][col] """
         """ A check if a queen can be placed on board[row][col].
         We just need to check ld[row-col+n-1] and rd[row+coln] 
         where ld and rd are for left and right diagonal respectively"""
         if ((ld[i - col + N - 1 ] ! = 1 and 
              rd[i + col] ! = 1 ) and cl[i] ! = 1 ):
                   
             """ Place this queen in board[i][col] """
             board[i][col] = 1
             ld[i - col + N - 1 ] = rd[i + col] = cl[i] = 1
              
             """ recur to place rest of the queens """
             if (solveNQUtil(board, col + 1 )):
                 return True
                  
             """ If placing queen in board[i][col] 
             doesn't lead to a solution, then remove queen from board[i][col] """
             board[i][col] = 0 # BACKTRACK 
             ld[i - col + N - 1 ] = rd[i + col] = cl[i] = 0
              
             """ If the queen cannot be placed in
             any row in this colum col then return False """
     return False
      
""" This function solves the N Queen problem using 
Backtracking. It mainly uses solveNQUtil() to 
solve the problem. It returns False if queens 
cannot be placed, otherwise, return True and 
prints placement of queens in the form of 1s. 
Please note that there may be more than one 
solutions, this function prints one of the 
feasible solutions."""
def solveNQ():
     board = [[ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ], [ 0 , 0 , 0 , 0 ]]
     if (solveNQUtil(board, 0 ) = = False ):
         printf( "Solution does not exist" )
         return False
     printSolution(board)
     return True
      
# Driver Code
solveNQ() 
  
# This code is contributed by SHUBHAMSINGH10

C#

/* C# program to solve N Queen Problem 
using backtracking */
using System;
      
class GFG 
{
static int N = 4;
  
/* ld is an array where its indices indicate row-col+N-1
(N-1) is for shifting the difference to store negative 
indices */
static int []ld = new int [30];
  
/* rd is an array where its indices indicate row+col
and used to check whether a queen can be placed on 
right diagonal or not*/
static int []rd = new int [30];
  
/*column array where its indices indicates column and 
used to check whether a queen can be placed in that
     row or not*/
static int []cl = new int [30];
  
/* A utility function to print solution */
static void printSolution( int [, ]board)
{
     for ( int i = 0; i < N; i++)
     {
         for ( int j = 0; j < N; j++)
             Console.Write( " {0} " , board[i, j]);
         Console.Write( "\n" );
     }
}
  
/* A recursive utility function to solve N
Queen problem */
static bool solveNQUtil( int [, ]board, int col)
{
     /* base case: If all queens are placed
     then return true */
     if (col >= N)
         return true ;
  
     /* Consider this column and try placing
     this queen in all rows one by one */
     for ( int i = 0; i < N; i++)
     {
          
         /* Check if the queen can be placed on
         board[i, col] */
         /* A check if a queen can be placed on 
         board[row, col].We just need to check
         ld[row-col+n-1] and rd[row+coln] where
         ld and rd are for left and right 
         diagonal respectively*/
         if ((ld[i - col + N - 1] != 1 &&
              rd[i + col] != 1) && cl[i] != 1)
         {
             /* Place this queen in board[i, col] */
             board[i, col] = 1;
             ld[i - col + N - 1] =
             rd[i + col] = cl[i] = 1;
  
             /* recur to place rest of the queens */
             if (solveNQUtil(board, col + 1))
                 return true ;
  
             /* If placing queen in board[i, col]
             doesn't lead to a solution, then
             remove queen from board[i, col] */
             board[i, col] = 0; // BACKTRACK
             ld[i - col + N - 1] =
             rd[i + col] = cl[i] = 0;
         }
     }
  
     /* If the queen cannot be placed in any row in
         this colum col then return false */
     return false ;
}
  
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to 
solve the problem. It returns false if queens
cannot be placed, otherwise, return true and
prints placement of queens in the form of 1s.
Please note that there may be more than one
solutions, this function prints one of the
feasible solutions.*/
static bool solveNQ()
{
     int [, ]board = {{ 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 0, 0 }};
  
     if (solveNQUtil(board, 0) == false ) 
     {
         Console.Write( "Solution does not exist" );
         return false ;
     }
  
     printSolution(board);
     return true ;
}
  
// Driver Code
public static void Main(String[] args)
{
     solveNQ();
}
}
  
// This code is contributed by Rajput-Ji

输出如下:

1值表示皇后区的位置

0  0  1  0 
 1  0  0  0 
 0  0  0  1 
 0  1  0  0

打印N皇后问题中的所有解决方案

资料来源:

http://see.stanford.edu/materials/icspacs106b/H19-RecBacktrackExamples.pdf

http://en.literateprograms.org/Eight_queens_puzzle_%28C%29

http://en.wikipedia.org/wiki/Eight_queens_puzzle

如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

木子山

发表评论

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen: