高级算法设计（打印N皇后问题中的所有解决方案） _N皇后问题所有解决方案

本文概述

C ++
Java
Python3
C#
C ++
Java
Python3

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 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问题的两个解决方案。

文章图片

文章图片
In
以前
帖子中, 我们讨论了一种仅打印一种可能的解决方案的方法, 因此, 现在本文中的任务是打印N-Queen问题中的所有解决方案。这里讨论的解决方案是相同方法的扩展。
推荐：请尝试以下方法
{IDE}
首先, 在继续解决方案之前。
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 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.

在代码中突出显示的上述算法只有一点修改。
C ++

/* C/C++ program to solve N Queen Problem using backtracking */ #include< bits/stdc++.h> #define N 4 /* A utility function to print solution */ void printSolution( int board[N][N]) { static int k = 1; printf ( "%d-\n" , k++); for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) printf ( " %d " , board[i][j]); printf ( "\n" ); } 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) { printSolution(board); return true ; } /* Consider this column and try placing this queen in all rows one by one */ bool res = false ; for ( int i = 0; i < N; i++) { /* Check if queen can be placed on board[i][col] */ if ( isSafe(board, i, col) ) { /* Place this queen in board[i][col] */ board[i][col] = 1; // Make result true if any placement // is possible res = solveNQUtil(board, col + 1) || res; /* 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 queen can not be place in any row in this column col then return false */ return res; } /* 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.*/ void solveNQ() { int board[N][N]; memset (board, 0, sizeof (board)); if (solveNQUtil(board, 0) == false ) { printf ( "Solution does not exist" ); return ; } return ; } // driver program to test above function int main() { solveNQ(); return 0; }

Java

/* Java program to solve N Queen Problem using backtracking */ class GfG { static int N = 4 ; static int k = 1 ; /* A utility function to print solution */ static void printSolution( int board[][]) { System.out.printf( "%d-\n" , k++); 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" ); } System.out.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 */ static 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 */ static boolean solveNQUtil( int board[][], int col) { /* base case: If all queens are placed then return true */ if (col == N) { printSolution(board); return true ; } /* Consider this column and try placing this queen in all rows one by one */ boolean res = false ; for ( int i = 0 ; i < N; i++) { /* Check if queen can be placed on board[i][col] */ if ( isSafe(board, i, col) ) { /* Place this queen in board[i][col] */ board[i][col] = 1 ; // Make result true if any placement // is possible res = solveNQUtil(board, col + 1 ) || res; /* 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 queen can not be place in any row in this column col then return false */ return res; } /* 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 void solveNQ() { int board[][] = new int [N][N]; if (solveNQUtil(board, 0 ) == false ) { System.out.printf( "Solution does not exist" ); return ; } return ; } // Driver code public static void main(String[] args) { solveNQ(); } } // This code has been contributed // by 29AjayKumar

Python3

''' Python3 program to solve N Queen Problem using backtracking ''' k = 1 # A utility function to print solution def printSolution(board): global k print (k, "-\n" ) k = k + 1 for i in range ( 4 ): for j in range ( 4 ): print (board[i][j], end = " " ) print ( "\n" ) print ( "\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 ''' def isSafe(board, row, col) :# Check this row on left side for i in range (col): if (board[row][i]): return False # Check upper diagonal on left side i = row j = col while i > = 0 and j > = 0 : if (board[i][j]): return False ; i - = 1 j - = 1 # Check lower diagonal on left side i = row j = col while j > = 0 and i < 4 : if (board[i][j]): return False i = i + 1 j = j - 1 return True ''' 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 = = 4 ): printSolution(board) return True ''' Consider this column and try placing this queen in all rows one by one ''' res = False for i in range ( 4 ):''' Check if queen can be placed on board[i][col] ''' if (isSafe(board, i, col)):# Place this queen in board[i][col] board[i][col] = 1 ; # Make result true if any placement # is possible res = solveNQUtil(board, col + 1 ) or res; ''' 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 queen can not be place in any row in this column col then return false ''' return res ''' 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 for j in range ( 10 )] for i in range ( 10 )] if (solveNQUtil(board, 0 ) = = False ):print ( "Solution does not exist" ) return return # Driver Code solveNQ() # This code is contributed by YatinGupta

/* C# program to solve N Queen Problem using backtracking */ using System; class GfG { static int N = 4; static int k = 1; /* A utility function to print solution */ static void printSolution( int [, ]board) { Console.Write( "{0}-\n" , k++); for ( int i = 0; i < N; i++) { for ( int j = 0; j < N; j++) Console.Write( " {0} " , board[i, j]); Console.Write( "\n" ); } Console.Write( "\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 */ static 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 */ static bool solveNQUtil( int [, ]board, int col) { /* base case: If all queens are placed then return true */ if (col == N) { printSolution(board); return true ; } /* Consider this column and try placing this queen in all rows one by one */ bool res = false ; for ( int i = 0; i < N; i++) { /* Check if queen can be placed on board[i, col] */ if ( isSafe(board, i, col) ) { /* Place this queen in board[i, col] */ board[i, col] = 1; // Make result true if any placement // is possible res = solveNQUtil(board, col + 1) || res; /* 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 queen can not be place in any row in this column col then return false */ return res; } /* 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 void solveNQ() { int [, ]board = new int [N, N]; if (solveNQUtil(board, 0) == false ) { Console.Write( "Solution does not exist" ); return ; } return ; } // Driver code public static void Main() { solveNQ(); } } /* This code contributed by PrinciRaj1992 */

输出如下

1- 0010 1000 0001 0100 2- 0100 0001 1000 0010

输出如下：

1- 0010 1000 0001 0100 2- 0100 0001 1000 0010

C ++

// CPP program for above approach #include < iostream> #include < vector> #include < math.h> using namespace std; // Program to print board void printBoard(vector< vector< char > > & board) { for ( auto row : board) { cout< < "|" ; for ( auto ch : row) { cout< < ch< < "|" ; } cout< < endl; } } // Program to solve N Quuens problem void solveBoard(vector< vector< char > > & board, int row, int rowmask, int ldmask, int rdmask, int & ways) { int n = board.size(); // All_rows_filled is a bit mask having all N bits set int all_rows_filled = (1 < < n) - 1; // If rowmask will have all bits set, means queen has been // placed successfully in all rows and board is diplayed if (rowmask == all_rows_filled) { ways++; cout< < "=====================\n" ; cout< < "Queen placement - " < < ways< < endl; cout< < "=====================\n" ; printBoard(board); return ; } // We extract a bit mask(safe) by rowmask, // ldmask and rdmask. all set bits of 'safe' // indicates the safe column index for queen // placement of this iteration for row index(row). int safe = all_rows_filled & (~(rowmask | ldmask | rdmask)); while (safe) { // Extracts the right-most set bit // (safe column index) where queen // can be placed for this row int p = safe & (-safe); int col = ( int )log2(p); board[row][col] = 'Q' ; // This is very important: // we need to update rowmask, ldmask and rdmask // for next row as safe index for queen placement // will be decided by these three bit masks. // We have all three rowmask, ldmask and // rdmask as 0 in beginning. Suppose, we are placing // queen at 1st column index at 0th row. rowmask, ldmask // and rdmask will change for next row as below: // rowmask's 1st bit will be set by OR operation // rowmask = 00000000000000000000000000000010 // ldmask will change by setting 1st // bit by OR operationand left shifting // by 1 as it has to block the next column // of next row because that will fall on left diagonal. // ldmask = 00000000000000000000000000000100 // rdmask will change by setting 1st bit // by OR operation and right shifting by 1 // as it has to block the previous column // of next row because that will fall on right diagonal. // rdmask = 00000000000000000000000000000001 // these bit masks will keep updated in each // iteration for next row solveBoard(board, row+1, rowmask|p, (ldmask|p) < < 1, (rdmask|p) > > 1, ways); // Reset right-most set bit to 0 so, // next iteration will continue by placing the queen // at another safe column index of this row safe = safe & (safe-1); // Backtracking, replace 'Q' by ' ' board[row][col] = ' ' ; } return ; } // Driver Code int main() { // Board size int n = 4; int ways = 0; vector< vector< char > > board; for ( int i = 0; i < n; i++) { vector< char > tmp; for ( int j = 0; j < n; j++) { tmp.push_back( ' ' ); } board.push_back(tmp); } int rowmask = 0, ldmask = 0, rdmask = 0; int row = 0; // Function Call solveBoard(board, row, rowmask, ldmask, rdmask, ways); cout< < endl< < "Number of ways to place " < < n< < " queens : " < < ways< < endl; return 0; } // This code is contributed by Nikhil Vinay

Java

// Java Program for aove approach public class NQueenSolution { static private int ways = 0 ; // Program to solve N-Queens Problem public void solveBoard( char [][] board, int row, int rowmask, int ldmask, int rdmask) { int n = board.length; // All_rows_filled is a bit mask // having all N bits set int all_rows_filled = ( 1 < < n) - 1 ; // If rowmask will have all bits set, // means queen has been placed successfully // in all rows and board is diplayed if (rowmask == all_rows_filled) { ways++; System.out.println( "=====================" ); System.out.println( "Queen placement - " + ways); System.out.println( "=====================" ); printBoard(board); }// We extract a bit mask(safe) by rowmask, // ldmask and rdmask. all set bits of 'safe' // indicates the safe column index for queen // placement of this iteration for row index(row). int safe = all_rows_filled & (~(rowmask | ldmask | rdmask)); while (safe > 0 ) {// Extracts the right-most set bit // (safe column index) where queen // can be placed for this row int p = safe & (-safe); int col = ( int )(Math.log(p) / Math.log( 2 )); board[row][col] = 'Q' ; // This is very important: // we need to update rowmask, ldmask and rdmask // for next row as safe index for queen placement // will be decided by these three bit masks. // We have all three rowmask, ldmask and // rdmask as 0 in beginning. Suppose, we are placing // queen at 1st column index at 0th row. rowmask, ldmask // and rdmask will change for next row as below:// rowmask's 1st bit will be set by OR operation // rowmask = 00000000000000000000000000000010 // ldmask will change by setting 1st // bit by OR operationand left shifting // by 1 as it has to block the next column // of next row because that will fall on left diagonal. // ldmask = 00000000000000000000000000000100 // rdmask will change by setting 1st bit // by OR operation and right shifting by 1 // as it has to block the previous column // of next row because that will fall on right diagonal. // rdmask = 00000000000000000000000000000001 // these bit masks will keep updated in each // iteration for next row solveBoard(board, row+ 1 , rowmask|p, (ldmask|p) < < 1 , (rdmask|p) > > 1 ); // Reset right-most set bit to 0 so, // next iteration will continue by placing the queen // at another safe column index of this row safe = safe & (safe- 1 ); // Backtracking, replace 'Q' by ' ' board[row][col] = ' ' ; } } // Program to print board public void printBoard( char [][] board) { for ( int i = 0 ; i < board.length; i++) { System.out.print( "|" ); for ( int j = 0 ; j < board[i].length; j++) { System.out.print(board[i][j] + "|" ); } System.out.println(); } } // Driver Code public static void main(String args[]) {// Board size int n = 4 ; char board[][] = new char [n][n]; for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { board[i][j] = ' ' ; } } int rowmask = 0 , ldmask = 0 , rdmask = 0 ; int row = 0 ; NQueenSolution solution = new NQueenSolution(); // Function Call solution.solveBoard(board, row, rowmask, ldmask, rdmask); System.out.println(); System.out.println( "Number of ways to place " + n + " queens : " + ways); } } // This code is contributed by Nikhil Vinay

Python3

# Python program for above approach import math ways = 0 # Program to solve N-Queens Problem def solveBoard(board, row, rowmask, ldmask, rdmask):n = len (board)# All_rows_filled is a bit mask # having all N bits set all_rows_filled = ( 1 < < n) - 1# If rowmask will have all bits set, means # queen has been placed successfully # in all rows and board is diplayed if (rowmask = = all_rows_filled): global ways ways = ways + 1 print ( "=====================" ) print ( "Queen placement - " + ( str )(ways)) print ( "=====================" ) printBoard(board)# We extract a bit mask(safe) by rowmask, # ldmask and rdmask. all set bits of 'safe' # indicates the safe column index for queen # placement of this iteration for row index(row). safe = all_rows_filled & (~(rowmask | ldmask | rdmask))while (safe > 0 ):# Extracts the right-most set bit # (safe column index) where queen # can be placed for this row p = safe & ( - safe) col = ( int )(math.log(p) / math.log( 2 )) board[row][col] = 'Q'# This is very important: # we need to update rowmask, ldmask and rdmask # for next row as safe index for queen placement # will be decided by these three bit masks. # We have all three rowmask, ldmask and # rdmask as 0 in beginning. Suppose, we are placing # queen at 1st column index at 0th row. rowmask, ldmask # and rdmask will change for next row as below:# rowmask's 1st bit will be set by OR operation # rowmask = 00000000000000000000000000000010 # ldmask will change by setting 1st # bit by OR operationand left shifting # by 1 as it has to block the next column # of next row because that will fall on left diagonal. # ldmask = 00000000000000000000000000000100 # rdmask will change by setting 1st bit # by OR operation and right shifting by 1 # as it has to block the previous column # of next row because that will fall on right diagonal. # rdmask = 00000000000000000000000000000001 # these bit masks will keep updated in each # iteration for next row solveBoard(board, row + 1 , rowmask|p, (ldmask|p) < < 1 , (rdmask|p) > > 1 )# Reset right-most set bit to 0 so, next # iteration will continue by placing the queen # at another safe column index of this row safe = safe & (safe - 1 )# Backtracking, replace 'Q' by ' ' board[row][col] = ' ' # Program to print board def printBoard(board): for row in board: print ( "|" + "|" .join(row) + "|" ) # Driver Code def main():n = 4 ; # board size board = []for i in range (n): row = [] for j in range (n): row.append( ' ' ) board.append(row) rowmask = 0 ldmask = 0 rdmask = 0 row = 0# Function Call solveBoard(board, row, rowmask, ldmask, rdmask)# creates a new line print () print ( "Number of ways to place " + ( str )(n) + " queens : " + ( str )(ways))if __name__ = = "__main__" : main() #This code is contributed by Nikhil Vinay

输出如下

===================== Queen placement - 1 ===================== | |Q| | | | | | |Q| |Q| | | | | | |Q| | ===================== Queen placement - 2 ===================== | | |Q| | |Q| | | | | | | |Q| | |Q| | |Number of ways to place 4 queens : 2

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