La Reina N es el problema de colocar N reinas de ajedrez en un tablero de ajedrez N × N para que no haya dos reinas que se ataquen entre sí. Por ejemplo, la siguiente es una solución para el problema de 4 Queen.
En publicaciones anteriores , hemos discutido un enfoque que imprime solo una solución posible, por lo que ahora en esta publicación, la tarea es imprimir todas las soluciones en N-Queen Problem. Cada solución contiene configuraciones de tablero distintas de la colocación de las N-reinas, donde las soluciones son una permutación de [1,2,3..n] en orden creciente, aquí el número en el i-ésimo lugar indica que la reina de la i-ésima columna es colocado en la fila con ese número. Para el ejemplo anterior, la solución se escribe como [[2 4 1 3 ] [3 1 4 2 ]]. La solución discutida aquí es una extensión del mismo enfoque.
Algoritmo de retroceso
La idea es colocar las reinas una por una en diferentes columnas, comenzando desde la columna más a la izquierda. Cuando colocamos una reina en una columna, buscamos conflictos con reinas ya colocadas. En la columna actual, si encontramos una fila para la que no hay conflicto, marcamos esta fila y columna como parte de la solución. Si no encontramos esa fila debido a conflictos, retrocedemos y devolvemos falso.
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.
Una modificación es que podemos encontrar si tenemos una reina previamente colocada en una columna o en diagonal izquierda o en diagonal derecha en tiempo O(1). Podemos observar que
- Para todas las celdas en una diagonal izquierda particular, su fila + columna = constante.
- Para todas las celdas en una diagonal derecha particular, su fila – col + n – 1 = constante.
Digamos que n = 5, entonces tenemos un total de 2n-1 diagonales izquierda y derecha
Digamos que colocamos una reina en (2,0)
(2,0) tienen el valor de la Diagonal izquierda = 2. Ahora no podemos colocar otra reina en (1,1) y (0,2) porque ambos tienen el mismo valor de la Diagonal izquierda que para (2,0). También se puede notar algo similar para la diagonal derecha.
Implementación:
C++
/* C/C++ program to solve N Queen Problem using
backtracking */
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > result;
/* 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(vector<vector<int> > board,
int row, int col)
{
int i, j;
int N = board.size();
/* 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(vector<vector<int> >& board, int col)
{
/* base case: If all queens are placed
then return true */
int N = board.size();
if (col == N) {
vector<int> v;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (board[i][j] == 1)
v.push_back(j + 1);
}
}
result.push_back(v);
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.*/
vector<vector<int> > nQueen(int n)
{
result.clear();
vector<vector<int> > board(n, vector<int>(n, 0));
if (solveNQUtil(board, 0) == false) {
return {};
}
sort(result.begin(), result.end());
return result;
}
// Driver Code
int main()
{
int n = 4;
vector<vector<int> > v = nQueen(n);
for (auto ar : v) {
cout << "[";
for (auto it : ar)
cout << it << " ";
cout << "]";
}
return 0;
}
Java
/* Java program to solve N Queen
Problem using backtracking */
import java.util.*;
class GfG {
/* This function solves the N Queen problem using
Backtracking. It mainly uses solveNQUtil() to
solve the problem.
*/
static List<List<Integer>> nQueen(int n) {
// cols[i] = true if there is a queen previously placed at ith column
cols = new boolean[n];
// leftDiagonal[i] = true if there is a queen previously placed at
// i = (row + col )th left diagonal
leftDiagonal = new boolean[2*n];
// rightDiagonal[i] = true if there is a queen previously placed at
// i = (row - col + n - 1)th rightDiagonal diagonal
rightDiagonal = new boolean[2*n];
result = new ArrayList<>();
List<Integer> temp = new ArrayList<>();
for(int i=0;i<n;i++)temp.add(0);
solveNQUtil(result,n,0,temp);
return result;
}
private static void solveNQUtil(List<List<Integer>> result,int n,int row,List<Integer> comb){
if(row==n){
// if row==n it means we have successfully placed all n queens.
// hence add current arrangement to our answer
// comb represent current combination
result.add(new ArrayList<>(comb));
return;
}
for(int col = 0;col<n;col++){
// if we have a queen previously placed in the current column
// or in current left or right diagonal we continue
if(cols[col] || leftDiagonal[row+col] || rightDiagonal[row-col+n])
continue;
// otherwise we place a queen at cell[row][col] and
//make current column, left diagonal and right diagonal true
cols[col] = leftDiagonal[row+col] = rightDiagonal[row-col+n] = true;
comb.set(col,row+1);
// then we goto next row
solveNQUtil(result,n,row+1,comb);
// then we backtrack and remove our currently placed queen
cols[col] = leftDiagonal[row+col] = rightDiagonal[row-col+n] = false;
}
}
static List<List<Integer> > result
= new ArrayList<List<Integer> >();
static boolean[] cols,leftDiagonal,rightDiagonal;
// Driver code
public static void main(String[] args)
{
int n = 4;
List<List<Integer> > res = nQueen(n);
System.out.println(res);
}
}
Python3
''' Python3 program to solve N Queen Problem using backtracking ''' result = [] # A utility function to print solution ''' 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): v = [] for i in board: for j in range(len(i)): if i[j] == 1: v.append(j+1) result.append(v) 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(n): result.clear() board = [[0 for j in range(n)] for i in range(n)] solveNQUtil(board, 0) result.sort() return result # Driver Code n = 4 res = solveNQ(n) print(res) # This code is contributed by YatinGupta
C#
/* C# program to solve N Queen
Problem using backtracking */
using System;
using System.Collections;
using System.Collections.Generic;
class GfG {
static List<List<int> > result = new List<List<int> >();
/* 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 N)
{
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, int N)
{
/* base case: If all queens are placed
then return true */
if (col == N) {
List<int> v = new List<int>();
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++) {
if (board[i, j] == 1)
v.Add(j + 1);
}
result.Add(v);
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, N)) {
/* Place this queen in board[i,col] */
board[i, col] = 1;
// Make result true if any placement
// is possible
res = solveNQUtil(board, col + 1, N) || 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 List<List<int> > solveNQ(int n)
{
result.Clear();
int[, ] board = new int[n, n];
solveNQUtil(board, 0, n);
return result;
}
// Driver code
public static void Main()
{
int n = 4;
List<List<int> > res = solveNQ(n);
for (int i = 0; i < res.Count; i++) {
Console.Write("[");
for (int j = 0; j < res[i].Count; j++) {
Console.Write(res[i][j]+ " ");
}
Console.Write("]");
}
}
}
/* This code contributed by PrinciRaj1992 */
Producción
[2 4 1 3 ][3 1 4 2 ]
Enfoque de retroceso eficiente mediante enmascaramiento de bits
Algoritmo:
siempre hay solo una reina en cada fila y cada columna, por lo que la idea de retroceder es comenzar a colocar la reina desde la columna más a la izquierda de cada fila y encontrar una columna donde la reina pueda colocarse sin colisión con las reinas colocadas anteriormente. Se repite desde la primera fila hasta la última fila. Al colocar una reina, se realiza un seguimiento como si no estuviera colisionando (en filas, columnas y diagonalmente) con reinas colocadas en filas anteriores. Una vez que se encuentra que la reina no se puede colocar en un índice de columna en particular en una fila, el algoritmo retrocede y cambia la posición de la reina colocada en la fila anterior, luego avanza para colocar la reina en la fila siguiente.
- Comience con el vector de tres bits que se usa para rastrear el lugar seguro para la ubicación de la reina en filas, columnas y diagonalmente en cada iteración.
- El vector de tres bits contendrá información de la siguiente manera:
- máscara de fila: establecer el índice de bits (i) de este vector de bits indicará que la reina no se puede colocar en la i-ésima columna de la siguiente fila.
- ldmask: establecer el índice de bits (i) de este vector de bits indicará que la reina no puede colocarse en la i-ésima columna de la siguiente fila. Representa el índice de columna inseguro para la siguiente fila que cae debajo de la diagonal izquierda de las reinas colocadas en las filas anteriores.
- rdmask: establecer el índice de bits (i) de este vector de bits indicará que la reina no se puede colocar en la i-ésima columna de la siguiente fila. Representa el índice de la columna insegura para la siguiente fila que cae en la diagonal derecha de las reinas colocadas en las filas anteriores.
- Hay una array (tablero) 2-D (NxN), que tendrá el carácter ‘ ‘ en todos los índices al principio y se llena con ‘Q’ fila por fila. Una vez que todas las filas se llenan con ‘Q’, la solución actual se inserta en la lista de resultados.
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program for above approach
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > result;
// Program to solve N Queens 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
// displayed
if (rowmask == all_rows_filled) {
vector<int> v;
for (int i = 0; i < board.size(); i++) {
for (int j = 0; j < board.size(); j++) {
if (board[i][j] == 'Q')
v.push_back(j + 1);
}
}
result.push_back(v);
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 operation and 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
result.clear();
solveBoard(board, row, rowmask, ldmask, rdmask, ways);
sort(result.begin(),result.end());
for (auto ar : result) {
cout << "[";
for (auto it : ar)
cout << it << " ";
cout << "]";
}
return 0;
}
// This code is contributed by Nikhil Vinay
Java
// Java Program for above approach
import java.util.*;
public class NQueenSolution {
static List<List<Integer> > result
= new ArrayList<List<Integer> >();
// 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 displayed
if (rowmask == all_rows_filled) {
List<Integer> v = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == 'Q')
v.add(j + 1);
}
}
result.add(v);
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 > 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 operation and 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
result.clear();
solution.solveBoard(board, row, rowmask, ldmask,
rdmask);
System.out.println(result);
}
}
// This code is contributed by Nikhil Vinay
Python3
# Python program for above approach
import math
result = []
# 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 displayed
if (rowmask == all_rows_filled):
v = []
for i in board:
for j in range(len(i)):
if i[j] == 'Q':
v.append(j+1)
result.append(v)
# 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 operation and 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
result.clear()
solveBoard(board, row, rowmask, ldmask, rdmask)
result.sort()
print(result)
if __name__ == "__main__":
main()
# This code is contributed by Nikhil Vinay
C#
// C# program to print all solutions in N-Queen Problem
using System;
using System.Collections;
using System.Collections.Generic;
public class NQueenSolution{
static List<List<int>> result = new List<List<int>>();
public void solveBoard(char[,] board, int row,int rowmask, int ldmask,int rdmask)
{
int n = board.GetLength(0);
// 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 displayed
if (rowmask == all_rows_filled) {
List<int> v = new List<int>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (board[i,j] == 'Q')
v.Add(j + 1);
}
}
result.Add(v);
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 > 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 operation and 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.GetLength(0); i++) {
Console.Write("|");
for (int j = 0; j < board.GetLength(1); j++) {
Console.Write(board[i,j] + "|");
}
Console.Write("\n");
}
}
static public void Main (){
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
result.Clear();
solution.solveBoard(board, row, rowmask, ldmask,rdmask);
foreach (var sublist in result)
{
Console.Write("[");
foreach (int item in sublist)
{
Console.Write(item+" ");
}
Console.Write("]");
}
}
}
// This code is contributed by shruti456rawal
Producción
[2 4 1 3 ][3 1 4 2 ]
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA