# 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 388 次浏览

## 本文概述

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。

``````{ 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。

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

``````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.``````

## 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
{

/* 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``````

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

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

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。

## 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 = { 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 = { 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 = { 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 ;

/* 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 ;

/*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 ;

/* 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``````

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``````

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 