考虑一下 N×N 棋盘上有王后和王后 K 障碍。女王不能通过障碍物。给定皇后的位置（x，y），任务是找到皇后可以移动的细胞数量。
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例如：
Input : N = 8, x = 4, y = 4, K = 0Output : 27
![图片[1]-棋盘上有障碍物时，女王可以移动的单元格数-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/geeks_chessboard-queen.png)
Input : N = 8, x = 4, y = 4, K = 1, kx1 = 3, ky1 = 5Output : 24
![图片[2]-棋盘上有障碍物时，女王可以移动的单元格数-yiteyi-C++库](https://www.yiteyi.com/wp-content/uploads/geeks/geeks_chessboard-queen-obstacle.png)
方法1： 这个想法是迭代女王可以攻击和停止的细胞，直到有障碍物或电路板的末端。要做到这一点，我们需要水平、垂直和对角地迭代。从位置（x，y）的移动可以是： （x+1，y）：向右水平移动一步。 （x-1，y）：向左水平移动一步。 （x+1，y+1）：一步对角向右移动。 （x-1，y-1）：一步斜移左下。 （x-1，y+1）：一步斜向上向左移动。 （x+1，y-1）：一步对角向下移动。 （x，y+1）：向下一步。 （x，y-1）：向上一步。
下面是C++实现这种方法：
C++
// C++ program to find number of cells a queen can move // with obstacles on the chessborad #include<bits/stdc++.h> using namespace std; // Return if position is valid on chessboard int range( int n, int x, int y) { return (x <= n && x > 0 && y <= n && y > 0); } // Return the number of moves with a given direction int check( int n, int x, int y, int xx, int yy, map <pair< int , int >, int > mp) { int ans = 0; // Checking valid move of Queen in a direction. while (range(n, x, y) && ! mp[{x, y}]) { x += xx; y += yy; ans++; } return ans; } // Return the number of position a Queen can move. int numberofPosition( int n, int k, int x, int y, int obstPosx[], int obstPosy[]) { int x1, y1, ans = 0; map <pair< int , int >, int > mp; // Mapping each obstacle's position while (k--) { x1 = obstPosx[k]; y1 = obstPosy[k]; mp[{x1, y1}] = 1; } // Fetching number of position a queen can // move in each direction. ans += check(n, x + 1, y, 1, 0, mp); ans += check(n, x-1, y, -1, 0, mp); ans += check(n, x, y + 1, 0, 1, mp); ans += check(n, x, y-1, 0, -1, mp); ans += check(n, x + 1, y + 1, 1, 1, mp); ans += check(n, x + 1, y-1, 1, -1, mp); ans += check(n, x-1, y + 1, -1, 1, mp); ans += check(n, x-1, y-1, -1, -1, mp); return ans; } // Driven Program int main() { int n = 8; // Chessboard size int k = 1; // Number of obstacles int Qposx = 4; // Queen x position int Qposy = 4; // Queen y position int obstPosx[] = { 3 }; // x position of obstacles int obstPosy[] = { 5 }; // y position of obstacles cout << numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy) << endl; return 0; } |
JAVA
// Java program to find number of cells a queen can move // with obstacles on the chessborad import java.util.*; class GFG{ static class pair { int first, second; public pair( int first, int second) { this .first = first; this .second = second; } } // Return if position is valid on chessboard static boolean range( int n, int x, int y) { return (x <= n && x > 0 && y <= n && y > 0 ); } // Return the number of moves with a given direction static int check( int n, int x, int y, int xx, int yy, HashMap <pair, Integer> mp) { int ans = 0 ; // Checking valid move of Queen in a direction. while (range(n, x, y) && ! mp.containsKey( new pair(x, y))) { x += xx; y += yy; ans++; } return ans; } // Return the number of position a Queen can move. static int numberofPosition( int n, int k, int x, int y, int obstPosx[], int obstPosy[]) { int x1, y1, ans = 0 ; HashMap <pair, Integer> mp = new HashMap<>(); // Mapping each obstacle's position while (k> 0 ) { k--; x1 = obstPosx[k]; y1 = obstPosy[k]; mp.put( new pair(x1, y1), 1 ); } // Fetching number of position a queen can // move in each direction. ans += check(n, x + 1 , y, 1 , 0 , mp); ans += check(n, x- 1 , y, - 1 , 0 , mp); ans += check(n, x, y + 1 , 0 , 1 , mp); ans += check(n, x, y- 1 , 0 , - 1 , mp); ans += check(n, x + 1 , y + 1 , 1 , 1 , mp); ans += check(n, x + 1 , y- 1 , 1 , - 1 , mp); ans += check(n, x- 1 , y + 1 , - 1 , 1 , mp); return ans; } // Driven Program public static void main(String[] args) { int n = 8 ; // Chessboard size int k = 1 ; // Number of obstacles int Qposx = 4 ; // Queen x position int Qposy = 4 ; // Queen y position int obstPosx[] = { 3 }; // x position of obstacles int obstPosy[] = { 5 }; // y position of obstacles System.out.print(numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy) + "" ); } } // This code contributed by Rajput-Ji |
C#
// C# program to find number of cells a queen can move // with obstacles on the chessborad using System; using System.Collections.Generic; public class GFG{ public class pair { public int first, second; public pair( int first, int second) { this .first = first; this .second = second; } } // Return if position is valid on chessboard static bool range( int n, int x, int y) { return (x <= n && x > 0 && y <= n && y > 0); } // Return the number of moves with a given direction static int check( int n, int x, int y, int xx, int yy, Dictionary <pair, int > mp) { int ans = 0; // Checking valid move of Queen in a direction. while (range(n, x, y) && ! mp.ContainsKey( new pair(x, y))) { x += xx; y += yy; ans++; } return ans; } // Return the number of position a Queen can move. static int numberofPosition( int n, int k, int x, int y, int []obstPosx, int []obstPosy) { int x1, y1, ans = 0; Dictionary <pair, int > mp = new Dictionary<pair, int >(); // Mapping each obstacle's position while (k>0) { k--; x1 = obstPosx[k]; y1 = obstPosy[k]; mp.Add( new pair(x1, y1), 1); } // Fetching number of position a queen can // move in each direction. ans += check(n, x + 1, y, 1, 0, mp); ans += check(n, x-1, y, -1, 0, mp); ans += check(n, x, y + 1, 0, 1, mp); ans += check(n, x, y-1, 0, -1, mp); ans += check(n, x + 1, y + 1, 1, 1, mp); ans += check(n, x + 1, y-1, 1, -1, mp); ans += check(n, x-1, y + 1, -1, 1, mp); return ans; } // Driven Program public static void Main(String[] args) { int n = 8; // Chessboard size int k = 1; // Number of obstacles int Qposx = 4; // Queen x position int Qposy = 4; // Queen y position int []obstPosx = { 3 }; // x position of obstacles int []obstPosy = { 5 }; // y position of obstacles Console.Write(numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy) + "" ); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // javascript program to find number of cells a queen can move // with obstacles on the chessborad class pair { constructor(first , second) { this .first = first; this .second = second; } } // Return if position is valid on chessboard function range(n , x , y) { return (x <= n && x > 0 && y <= n && y > 0); } // Return the number of moves with a given direction function check(n , x , y , xx , yy, mp) { var ans = 0; // Checking valid move of Queen in a direction. while (range(n, x, y) && !mp.has( new pair(x, y))) { x += xx; y += yy; ans++; } return ans; } // Return the number of position a Queen can move. function numberofPosition(n , k , x , y , obstPosx , obstPosy) { var x1, y1, ans = 0; var mp = new Map(); // Mapping each obstacle's position while (k > 0) { k--; x1 = obstPosx[k]; y1 = obstPosy[k]; mp.set( new pair(x1, y1), 1); } // Fetching number of position a queen can // move in each direction. ans += check(n, x + 1, y, 1, 0, mp); ans += check(n, x - 1, y, -1, 0, mp); ans += check(n, x, y + 1, 0, 1, mp); ans += check(n, x, y - 1, 0, -1, mp); ans += check(n, x + 1, y + 1, 1, 1, mp); ans += check(n, x + 1, y - 1, 1, -1, mp); ans += check(n, x - 1, y + 1, -1, 1, mp); return ans; } // Driven Program var n = 8; // Chessboard size var k = 1; // Number of obstacles var Qposx = 4; // Queen x position var Qposy = 4; // Queen y position var obstPosx = [ 3 ]; // x position of obstacles var obstPosy = [ 5 ]; // y position of obstacles document.write(numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy) + "" ); // This code is contributed by Rajput-Ji </script> |
输出：
24
方法2： 我们的想法是在障碍物上迭代，对于那些在女王的道路上的人，我们计算到达障碍物的自由细胞。如果路径中没有障碍物，我们必须计算沿着该方向到达电路板末端的自由单元数。
任何（x） 1. Y 1. )和（x） 2. Y 2. ):
- 如果它们水平处于同一水平：abs（x 1. –x 2. – 1)
- 如果它们垂直于同一水平面：abs（y 1. –y 2. –1）是之间的自由单元数。
- 如果是对角线：两个腹肌（x 1. –x 2. –1）或abs（y） 1. –y 2. –1）是之间的自由单元数。
以下是该方法的实施情况：
C++
// C++ program to find number of cells a queen can move // with obstacles on the chessborad #include <bits/stdc++.h> using namespace std; // Return the number of position a Queen can move. int numberofPosition( int n, int k, int x, int y, int obstPosx[], int obstPosy[]) { // d11, d12, d21, d22 are for diagonal distances. // r1, r2 are for vertical distance. // c1, c2 are for horizontal distance. int d11, d12, d21, d22, r1, r2, c1, c2; // Initialise the distance to end of the board. d11 = min( x-1, y-1 ); d12 = min( n-x, n-y ); d21 = min( n-x, y-1 ); d22 = min( x-1, n-y ); r1 = y-1; r2 = n-y; c1 = x-1; c2 = n-x; // For each obstacle find the minimum distance. // If obstacle is present in any direction, // distance will be updated. for ( int i = 0; i < k; i++) { if ( x > obstPosx[i] && y > obstPosy[i] && x-obstPosx[i] == y-obstPosy[i] ) d11 = min(d11, x-obstPosx[i]-1); if ( obstPosx[i] > x && obstPosy[i] > y && obstPosx[i]-x == obstPosy[i]-y ) d12 = min( d12, obstPosx[i]-x-1); if ( obstPosx[i] > x && y > obstPosy[i] && obstPosx[i]-x == y-obstPosy[i] ) d21 = min(d21, obstPosx[i]-x-1); if ( x > obstPosx[i] && obstPosy[i] > y && x-obstPosx[i] == obstPosy[i]-y ) d22 = min(d22, x-obstPosx[i]-1); if ( x == obstPosx[i] && obstPosy[i] < y ) r1 = min(r1, y-obstPosy[i]-1); if ( x == obstPosx[i] && obstPosy[i] > y ) r2 = min(r2, obstPosy[i]-y-1); if ( y == obstPosy[i] && obstPosx[i] < x ) c1 = min(c1, x-obstPosx[i]-1); if ( y == obstPosy[i] && obstPosx[i] > x ) c2 = min(c2, obstPosx[i]-x-1); } return d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; } // Driver code int main( void ) { int n = 8; // Chessboard size int k = 1; // number of obstacles int Qposx = 4; // Queen x position int Qposy = 4; // Queen y position int obstPosx[] = { 3 }; // x position of obstacles int obstPosy[] = { 5 }; // y position of obstacles cout << numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy); return 0; } |
JAVA
// Java program to find number of cells a // queen can move with obstacles on the // chessborad import java.io.*; class GFG { // Return the number of position a Queen // can move. static int numberofPosition( int n, int k, int x, int y, int obstPosx[], int obstPosy[]) { // d11, d12, d21, d22 are for diagonal distances. // r1, r2 are for vertical distance. // c1, c2 are for horizontal distance. int d11, d12, d21, d22, r1, r2, c1, c2; // Initialise the distance to end of the board. d11 = Math.min( x- 1 , y- 1 ); d12 = Math.min( n-x, n-y ); d21 = Math.min( n-x, y- 1 ); d22 = Math.min( x- 1 , n-y ); r1 = y- 1 ; r2 = n-y; c1 = x- 1 ; c2 = n-x; // For each obstacle find the minimum distance. // If obstacle is present in any direction, // distance will be updated. for ( int i = 0 ; i < k; i++) { if ( x > obstPosx[i] && y > obstPosy[i] && x-obstPosx[i] == y-obstPosy[i] ) d11 = Math.min(d11, x-obstPosx[i]- 1 ); if ( obstPosx[i] > x && obstPosy[i] > y && obstPosx[i]-x == obstPosy[i]-y ) d12 = Math.min( d12, obstPosx[i]-x- 1 ); if ( obstPosx[i] > x && y > obstPosy[i] && obstPosx[i]-x == y-obstPosy[i] ) d21 = Math.min(d21, obstPosx[i]-x- 1 ); if ( x > obstPosx[i] && obstPosy[i] > y && x-obstPosx[i] == obstPosy[i]-y ) d22 = Math.min(d22, x-obstPosx[i]- 1 ); if ( x == obstPosx[i] && obstPosy[i] < y ) r1 = Math.min(r1, y-obstPosy[i]- 1 ); if ( x == obstPosx[i] && obstPosy[i] > y ) r2 = Math.min(r2, obstPosy[i]-y- 1 ); if ( y == obstPosy[i] && obstPosx[i] < x ) c1 = Math.min(c1, x-obstPosx[i]- 1 ); if ( y == obstPosy[i] && obstPosx[i] > x ) c2 = Math.min(c2, obstPosx[i]-x- 1 ); } return d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; } // Driver code public static void main (String[] args) { int n = 8 ; // Chessboard size int k = 1 ; // number of obstacles int Qposx = 4 ; // Queen x position int Qposy = 4 ; // Queen y position int obstPosx[] = { 3 }; // x position of obstacles int obstPosy[] = { 5 }; // y position of obstacles System.out.println(numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy)); } } // This code is contributed by anuj_67. |
Python3
# Python program to find number of cells a # queen can move with obstacles on the # chessborad # Return the number of position a Queen # can move. def numberofPosition(n, k, x, y, obstPosx, obstPosy): # d11, d12, d21, d22 are for diagonal distances. # r1, r2 are for vertical distance. # c1, c2 are for horizontal distance. # Initialise the distance to end of the board. d11 = min (x - 1 , y - 1 ); d12 = min (n - x, n - y); d21 = min (n - x, y - 1 ); d22 = min (x - 1 , n - y); r1 = y - 1 ; r2 = n - y; c1 = x - 1 ; c2 = n - x; # For each obstacle find the minimum distance. # If obstacle is present in any direction, # distance will be updated. for i in range (k): if (x > obstPosx[i] and y > obstPosy[i] and x - obstPosx[i] = = y - obstPosy[i]): d11 = min (d11, x - obstPosx[i] - 1 ); if (obstPosx[i] > x and obstPosy[i] > y and obstPosx[i] - x = = obstPosy[i] - y): d12 = min (d12, obstPosx[i] - x - 1 ); if (obstPosx[i] > x and y > obstPosy[i] and obstPosx[i] - x = = y - obstPosy[i]): d21 = min (d21, obstPosx[i] - x - 1 ); if (x > obstPosx[i] and obstPosy[i] > y and x - obstPosx[i] = = obstPosy[i] - y): d22 = min (d22, x - obstPosx[i] - 1 ); if (x = = obstPosx[i] and obstPosy[i] < y): r1 = min (r1, y - obstPosy[i] - 1 ); if (x = = obstPosx[i] and obstPosy[i] > y): r2 = min (r2, obstPosy[i] - y - 1 ); if (y = = obstPosy[i] and obstPosx[i] < x): c1 = min (c1, x - obstPosx[i] - 1 ); if (y = = obstPosy[i] and obstPosx[i] > x): c2 = min (c2, obstPosx[i] - x - 1 ); return d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; # Driver code if __name__ = = '__main__' : n = 8 ; # Chessboard size k = 1 ; # number of obstacles Qposx = 4 ; # Queen x position Qposy = 4 ; # Queen y position obstPosx = [ 3 ] ; # x position of obstacles obstPosy = [ 5 ]; # y position of obstacles print (numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy)); # This code is contributed by Rajput-Ji |
C#
// C# program to find number of cells a // queen can move with obstacles on the // chessborad using System; class GFG { // Return the number of position a Queen // can move. static int numberofPosition( int n, int k, int x, int y, int []obstPosx, int []obstPosy) { // d11, d12, d21, d22 are for diagonal distances. // r1, r2 are for vertical distance. // c1, c2 are for horizontal distance. int d11, d12, d21, d22, r1, r2, c1, c2; // Initialise the distance to end of the board. d11 = Math.Min( x- 1 , y- 1 ); d12 = Math.Min( n-x, n-y ); d21 = Math.Min( n-x, y- 1 ); d22 = Math.Min( x- 1 , n-y ); r1 = y- 1 ; r2 = n-y; c1 = x- 1 ; c2 = n-x; // For each obstacle find the Minimum distance. // If obstacle is present in any direction, // distance will be updated. for ( int i = 0 ; i < k; i++) { if ( x > obstPosx[i] && y > obstPosy[i] && x-obstPosx[i] == y-obstPosy[i] ) d11 = Math.Min(d11, x-obstPosx[i]- 1 ); if ( obstPosx[i] > x && obstPosy[i] > y && obstPosx[i]-x == obstPosy[i]-y ) d12 = Math.Min( d12, obstPosx[i]-x- 1 ); if ( obstPosx[i] > x && y > obstPosy[i] && obstPosx[i]-x == y-obstPosy[i] ) d21 = Math.Min(d21, obstPosx[i]-x- 1 ); if ( x > obstPosx[i] && obstPosy[i] > y && x-obstPosx[i] == obstPosy[i]-y) d22 = Math.Min(d22, x-obstPosx[i]- 1 ); if ( x == obstPosx[i] && obstPosy[i] < y ) r1 = Math.Min(r1, y-obstPosy[i]- 1 ); if ( x == obstPosx[i] && obstPosy[i] > y ) r2 = Math.Min(r2, obstPosy[i]-y- 1 ); if ( y == obstPosy[i] && obstPosx[i] < x ) c1 = Math.Min(c1, x-obstPosx[i]- 1 ); if ( y == obstPosy[i] && obstPosx[i] > x ) c2 = Math.Min(c2, obstPosx[i]-x- 1 ); } return d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; } // Driver code public static void Main () { int n = 8 ; // Chessboard size int k = 1 ; // number of obstacles int Qposx = 4 ; // Queen x position int Qposy = 4 ; // Queen y position int []obstPosx = { 3 }; // x position of obstacles int []obstPosy = { 5 }; // y position of obstacles Console.WriteLine(numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy)); } } // This code is contributed by anuj_67. |
PHP
<?php //PHP program to find number of cells a queen can move // with obstacles on the chessborad // Return the number of position a Queen can move. function numberofPosition( $n , $k , $x , $y , $obstPosx , $obstPosy ) { // d11, d12, d21, d22 are for diagonal distances. // r1, r2 are for vertical distance. // c1, c2 are for horizontal distance. $d11 ; $d12 ; $d21 ; $d22 ; $r1 ; $r2 ; $c1 ; $c2 ; // Initialise the distance to end of the board. $d11 = min( $x -1, $y -1 ); $d12 = min( $n - $x , $n - $y ); $d21 = min( $n - $x , $y -1 ); $d22 = min( $x -1, $n - $y ); $r1 = $y -1; $r2 = $n - $y ; $c1 = $x -1; $c2 = $n - $x ; // For each obstacle find the minimum distance. // If obstacle is present in any direction, // distance will be updated. for ( $i = 0; $i < $k ; $i ++) { if ( $x > $obstPosx [ $i ] && $y > $obstPosy [ $i ] && $x - $obstPosx [ $i ] == $y - $obstPosy [ $i ] ) $d11 = min( $d11 , $x - $obstPosx [ $i ]-1); if ( $obstPosx [ $i ] > $x && $obstPosy [ $i ] > $y && $obstPosx [ $i ]- $x == $obstPosy [ $i ]- $y ) $d12 = min( $d12 , $obstPosx [ $i ]- $x -1); if ( $obstPosx [ $i ] > $x && $y > $obstPosy [ $i ] && $obstPosx [ $i ]- $x == $y - $obstPosy [ $i ] ) $d21 = min( $d21 , $obstPosx [ $i ]- $x -1); if ( $x > $obstPosx [ $i ] && $obstPosy [ $i ] > $y && $x - $obstPosx [ $i ] == $obstPosy [ $i ]- $y ) $d22 = min( $d22 , $x - $obstPosx [ $i ]-1); if ( $x == $obstPosx [ $i ] && $obstPosy [ $i ] < $y ) $r1 = min( $r1 , $y - $obstPosy [ $i ]-1); if ( $x == $obstPosx [ $i ] && $obstPosy [ $i ] > $y ) $r2 = min( $r2 , $obstPosy [ $i ]- $y -1); if ( $y == $obstPosy [ $i ] && $obstPosx [ $i ] < $x ) $c1 = min( $c1 , $x - $obstPosx [ $i ]-1); if ( $y == $obstPosy [ $i ] && $obstPosx [ $i ] > $x ) $c2 = min( $c2 , $obstPosx [ $i ]- $x -1); } return $d11 + $d12 + $d21 + $d22 + $r1 + $r2 + $c1 + $c2 ; } // Driver code $n = 8; // Chessboard size $k = 1; // number of obstacles $Qposx = 4; // Queen x position $Qposy = 4; // Queen y position $obstPosx = array (3 ); // x position of obstacles $obstPosy = array (5 ); // y position of obstacles echo numberofPosition( $n , $k , $Qposx , $Qposy , $obstPosx , $obstPosy ); // This code is contributed by ajit. ?> |
Javascript
<script> // Javascript program to find number of cells a queen // can move with obstacles on the chessborad // Return the number of position a Queen can move. function numberofPosition( n, k, x, y, obstPosx, obstPosy) { // d11, d12, d21, d22 are for diagonal distances. // r1, r2 are for vertical distance. // c1, c2 are for horizontal distance. let d11, d12, d21, d22, r1, r2, c1, c2; // Initialise the distance to end of the board. d11 = Math.min( x-1, y-1 ); d12 = Math.min( n-x, n-y ); d21 = Math.min( n-x, y-1 ); d22 = Math.min( x-1, n-y ); r1 = y-1; r2 = n-y; c1 = x-1; c2 = n-x; // For each obstacle find the minimum distance. // If obstacle is present in any direction, // distance will be updated. for (let i = 0; i < k; i++) { if ( x > obstPosx[i] && y > obstPosy[i] && x-obstPosx[i] == y-obstPosy[i] ) d11 = Math.min(d11, x-obstPosx[i]-1); if ( obstPosx[i] > x && obstPosy[i] > y && obstPosx[i]-x == obstPosy[i]-y ) d12 = Math.min( d12, obstPosx[i]-x-1); if ( obstPosx[i] > x && y > obstPosy[i] && obstPosx[i]-x == y-obstPosy[i] ) d21 = Math.min(d21, obstPosx[i]-x-1); if ( x > obstPosx[i] && obstPosy[i] > y && x-obstPosx[i] == obstPosy[i]-y ) d22 = Math.min(d22, x-obstPosx[i]-1); if ( x == obstPosx[i] && obstPosy[i] < y ) r1 = min(r1, y-obstPosy[i]-1); if ( x == obstPosx[i] && obstPosy[i] > y ) r2 = Math.min(r2, obstPosy[i]-y-1); if ( y == obstPosy[i] && obstPosx[i] < x ) c1 = Math.min(c1, x-obstPosx[i]-1); if ( y == obstPosy[i] && obstPosx[i] > x ) c2 = Math.min(c2, obstPosx[i]-x-1); } return d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; } // Driver Code let n = 8; // Chessboard size let k = 1; // number of obstacles let Qposx = 4; // Queen x position let Qposy = 4; // Queen y position let obstPosx = [ 3 ]; // x position of obstacles let obstPosy = [ 5 ]; // y position of obstacles document.write(numberofPosition(n, k, Qposx, Qposy, obstPosx, obstPosy)); </script> |
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