回溯算法（N皇后问题解析和多语言代码实现）

本文概述

建议：在继续解决方案之前, 请先在"实践"上解决它。
C / C ++
Java
Python3
C#
C / C ++
Java
Python3
C#

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。

文章图片
【回溯算法（N皇后问题解析和多语言代码实现）】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 configurationif 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 column2) If all queens are placedreturn true3) 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 placingqueen here leads to a solution.b) If placing the queen in [row, column] leads toa solution then return true.c) If placing queen doesn't lead to a solution thenunmark 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 colthen 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 oneof 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 = 4def 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 Falsereturn Truedef 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 FalseprintSolution(board) return True# Driver Code solveNQ()# This code is contributed by Divyanshu Mehta

// 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

0010 1000 0001 0100

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 colthen 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 oneof 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# 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

0010 1000 0001 0100

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
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