ñ* N皇后算法得到的坐标 [英] N * N queen algorithm getting the coordinates
问题描述
我试图实施 N *ñ
皇后算法有一点扭曲它。在这个版本中的女王还可以将喜欢的骑士......
一切工作正常,但我试图让所有可能的解决方案的坐标。问题是,如果我把它放在山坳==ñ
它只是打印最后一个。如何解决这个任何想法?
静态无效的位置(INT山坳,诠释女王[],INT N){
// INT溶液= 0;
对于(INT行= 1;行< = N;排++){
皇后[COL] =行;
如果((check_Queen(行,列,皇后))==真)
{
如果((check_KnightMove(行,列,皇后))==真)
{
如果(COL == N)
{
的System.out.println((+行+,+列);
的System.out.println(溶液=+溶液);
解决方案++;
}
其他
{
安置(COL + 1,王后,N);
}
}
}
}
皇后[COL] = 0;
}
公共静态无效的主要(字串[] args){
INT溶液= 0;
扫描仪扫描=新的扫描仪(System.in);
System.out.print(请输入N);
INT N = scanner.nextInt(); // TODO自动生成方法存根
INT皇后[] =新的INT [N + 1];
安置(1,王后,N);
的System.out.println(nQueens:液=+解决方案);
}
静态布尔check_Queen(INT行,诠释山坳,诠释女王[])
{
//布尔标志= FALSE;
的for(int i = 1; I<西;我++)
{
如果(皇后[COL-I] ==排||
皇后[COL-I] ==行我||
皇后[COL-I] ==行+ I){
//标志= FALSE;
返回false;
}
}
返回true;
}
静态布尔check_KnightMove(INT行,诠释山坳,诠释女王[])
{
如果(COL> = 2及及(皇后[COL-2] ==(行-1)||皇后[COL-2] ==(行+ 1)||皇后[COL-1] ==(行-2)||皇后[COL-1] ==(行+ 2)))
{
返回false;
}
返回true;
}
}
这是很难说什么是错的解决方案中的不知道如何 check_Queen
和 check_KnightMove
定义。下面是我会怎么解决的任务:
公共类皇后
{
静态无效printSolution(INT []皇后)
{
INT L = queens.length;
的for(int i = 0; I<升;我++)
{
对于(INT J = 0; J< L,J ++)
{
System.out.print(皇后[我] == J'Q':'。');
System.out.print('');
}
的System.out.println();
}
的System.out.println();
}
静态INT位置(INT []王后,INT C)
{
如果(C == queens.length)
{
printSolution(皇后);
返回1;
}
其他
{
INT solutionCount = 0;
INT L = queens.length;
对于(INT R = 0;为r,L,R ++)
{
布尔标志= FALSE;
的for(int i = 0;我c为C;我++)
{
INT XD = C - 我;
INT YD = Math.abs(R - 皇后[I]);
如果(YD == 0 || XD ==码)
{
标志=真正的;
打破;
}
//骑士举措支持
如果((XD == 1&安培;&安培;码== 2)||(XD == 2&安培;&安培;码== 1))
{
标志=真正的;
打破;
}
}
如果(!标志)
{
皇后[C] = R;
solutionCount + =安置区(Queens,C + 1);
}
}
返回solutionCount;
}
}
公共静态无效的主要(字串[] args)
{
的System.out.println(
一共找到解决方案:+就业(新INT [11],0));
}
}
在您的解决方案,方法 check_KnightMove
不正确。当山坳== 2
它不允许皇后被放置在第1行,因为它认为总有一个女王在(西:0 ,排:0)
,即外板。下面是修改后的版本:
静态布尔check_KnightMove(INT行,诠释山坳,诠释女王[])
{
如果(COL> = 3
&功放;&安培; (皇后[COL - 2] ==(行 - 1)||皇后[COL - 2] ==(行+ 1)))
{
返回false;
}
如果(COL> = 2
&功放;&安培; (皇后[COL - 1] ==(行 - 2)||皇后[COL - 1] ==(行+ 2)))
{
返回false;
}
返回true;
}
I'm trying to implement the N*N
queen algorithm with a little twist to it. In this version the queen can also move like a knight...
Everything is working fine, but I'm trying to get the coordinates of all the possible solutions. The problem is that if I put it inside col == n
it just prints the last one. Any ideas of how to solve this?
static void placement(int col, int queens[], int n){
//int solution =0;
for (int row = 1; row <= n; row++) {
queens[col] = row;
if((check_Queen(row,col,queens)) == true)
{
if((check_KnightMove(row,col,queens)) == true)
{
if(col == n)
{
System.out.println("("+row + "," + col);
System.out.println("solution=" + solution);
solution++;
}
else
{
placement(col+1,queens,n);
}
}
}
}
queens[col] = 0;
}
public static void main(String[] args) {
int solution =0;
Scanner scanner=new Scanner(System.in);
System.out.print("Please enter N");
int n = scanner.nextInt();// TODO Auto-generated method stub
int queens[] = new int[n+1];
placement(1,queens,n);
System.out.println("nQueens: solution=" + solution);
}
static boolean check_Queen(int row, int col, int queens[])
{
//boolean flag = false;
for(int i =1; i<col; i++)
{
if (queens[col-i] == row ||
queens[col-i] == row-i ||
queens[col-i] == row+i) {
//flag = false;
return false;
}
}
return true;
}
static boolean check_KnightMove(int row, int col, int queens[])
{
if(col>=2&&(queens[col-2] == (row -1) || queens[col-2] == (row+1) || queens[col-1] == (row-2) || queens[col-1] == (row+2)))
{
return false;
}
return true;
}
}
It is hard to tell what is wrong in your solution without knowing how check_Queen
and check_KnightMove
are defined. Here is how I would solve the task:
public class Queens
{
static void printSolution (int [] queens)
{
int l = queens.length;
for (int i = 0; i < l; i++)
{
for (int j = 0; j < l; j++)
{
System.out.print (queens [i] == j ? 'Q' : '.');
System.out.print (' ');
}
System.out.println ();
}
System.out.println ();
}
static int placement (int [] queens, int c)
{
if (c == queens.length)
{
printSolution (queens);
return 1;
}
else
{
int solutionCount = 0;
int l = queens.length;
for (int r = 0; r < l; r++)
{
boolean flag = false;
for (int i = 0; i < c; i++)
{
int xd = c - i;
int yd = Math.abs (r - queens [i]);
if (yd == 0 || xd == yd)
{
flag = true;
break;
}
// Knight move support
if ((xd == 1 && yd == 2) || (xd == 2 && yd == 1))
{
flag = true;
break;
}
}
if (!flag)
{
queens [c] = r;
solutionCount += placement (queens, c + 1);
}
}
return solutionCount;
}
}
public static void main (String [] args)
{
System.out.println (
"Total solutions found: " + placement (new int [11], 0));
}
}
In your solution, method check_KnightMove
is incorrect. When col == 2
it does not allow Queen to be placed at row 1, because it believes that there is always a Queen at (col: 0, row: 0)
, i.e. outside the board. Here is corrected version:
static boolean check_KnightMove (int row, int col, int queens[])
{
if (col >= 3
&& (queens [col - 2] == (row - 1) || queens [col - 2] == (row + 1)))
{
return false;
}
if (col >= 2
&& (queens [col - 1] == (row - 2) || queens [col - 1] == (row + 2)))
{
return false;
}
return true;
}
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