国际象棋游戏删除对角线移动，如果棋子为主教/王后阻挡(java)

Question

我正在用Java制作一款国际象棋游戏。

我做了一个JFrame，它可以让我创建棋子，这就是为什么我对任何棋子都有所有可能的移动(我将比普通国际象棋中的棋子更多(。

但我有一个小问题，我已经两天试图删除主教的运动了，这并不像看起来那么容易。

我有一个数组，里面包含的位置像这样：

_______ PIECES[x][y] //1 is black 0 is no piece 2 is white
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
2 2 2 2 2 2 2 2 
2 2 2 2 2 2 2 2 
_______ legalmoves[y][x] //Containing legal moves 1 is move/attack 0 cannot move there
// ( (x,y) (its y x in this tab)
// is reverse because it needs to be somewhere else in the code this is not a big deal)
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
1 0 0 0 0 0 0 0 
0 1 0 0 0 0 0 0 
0 0 1 0 0 0 0 0 
0 0 0 1 0 0 0 1 
0 0 0 0 1 0 1 0 
0 0 0 0 0 0 0 0 
My piece X,Y : 5 7

当函数需要知道棋子的合法移动并且检测到棋子有对角线移动时，就会调用该函数对于这篇文章，这里的法律动作应该看起来像这个

0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0

因为还有其他碎片(白色(阻挡的移动

提前谢谢你，如果你愿意(如果它能帮助你帮助我(，我可以给你提供rooks的动作功能(当检测到直线动作时，它也适用于女王和我创作的任何有这种动作的作品(。车代码：

boolean t = false;
if (hasRectMoves) {
System.out.println(x + " XY " + y);
System.out.println("HAS RECT MOVES !");
System.out.println("_______ PIECES");
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
System.out.print(pieces[i][j] + " ");
}
System.out.println("");
}
System.out.println("_______ RES");
for (int i = 0; i < 8; i++) {
//System.out.println("i = "+ i);
for (int j = 0; j < 8; j++) {
System.out.print(res[j][i] + " ");
}
System.out.println("");
}
System.out.println("My piece : " + (x + 1) + " " + (y + 1));
boolean test = true;
for (int i = y; i >= 0; i--) {
//res[x+1][i]=0;
if (test) {
if (pieces[i][x + 1] != 0) {
if (this.isBlack) {
if (pieces[i][x + 1] == 1)
res[x + 1][i] = 4;//4 to seen in the array where it makes move illegal
}
if (!this.isBlack) {
if (pieces[i][x + 1] == 2)
res[x + 1][i] = 4;
}
// System.out.println(i + " "+ pieces[i][y+1] + " "  + (x+1) + "=x y="+ (y+1)); 
test = false;
}
} else {
res[x + 1][i] = 4;
}
}
test = true;
for (int i = x; i >= 0; i--) {
//res[x+1][i]=0;
if (test) {
if (pieces[y + 1][i] != 0) {
if (this.isBlack) {
if (pieces[y + 1][i] == 1)
res[i][y + 1] = 4;
}
if (!this.isBlack) {
if (pieces[y + 1][i] == 2)
res[i][y + 1] = 4;
}
// System.out.println(i + " "+ pieces[i][y+1] + " "  + (x+1) + "=x y="+ (y+1)); 
test = false;
}
} else {
res[i][y + 1] = 4;
}
}
test = true;
for (int i = x + 2; i < 8; i++) {
//res[x+1][i]=0;
if (test) {
if (pieces[y + 1][i] != 0) {
if (this.isBlack) {
if (pieces[y + 1][i] == 1)
res[i][y + 1] = 4;
}
if (!this.isBlack) {
if (pieces[y + 1][i] == 2)
res[i][y + 1] = 4;
}
// System.out.println(i + " "+ pieces[i][y+1] + " "  + (x+1) + "=x y="+ (y+1)); 
test = false;
}
} else {
res[i][y + 1] = 4;
}
}
test = true;
for (int i = y + 2; i < 8; i++) {
//res[x+1][i]=0;
if (test) {
if (pieces[i][x + 1] != 0) {
if (this.isBlack) {
if (pieces[i][x + 1] == 1)
res[x + 1][i] = 4;
}
if (!this.isBlack) {
if (pieces[i][x + 1] == 2)
res[x + 1][i] = 4;
}
// System.out.println(i + " "+ pieces[i][y+1] + " "  + (x+1) + "=x y="+ (y+1)); 
test = false;
}
} else {
res[x + 1][i] = 4;
}
}

Answer 1

不错的项目。

我写了一个主要的小例子，所以你可以尝试一下。当然还有改进的余地。

我的输出：

3 XY 4
HAS RECT MOVES !
_______ PIECES
1 1 1 1 1 1 1 1 
1 1 1 1 1 1 1 1 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
2 2 2 2 2 2 2 2 
2 2 2 2 2 2 2 2 
_______ RES
0 0 0 0 0 0 0 5 
6 0 0 0 0 0 6 0 
0 7 0 0 0 7 0 0 
0 0 8 0 8 0 0 0 
0 0 0 9 0 0 0 0 
0 0 8 0 8 0 0 0 
0 7 0 0 0 7 0 0 
6 0 0 0 0 0 6 0 
My piece : 3 4
_______ legalmoves
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 
0 7 0 0 0 7 0 0 
0 0 8 0 8 0 0 0 
0 0 0 9 0 0 0 0 
0 0 8 0 8 0 0 0 
0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 

主教的职位标为9。我给每一个场地加上一个从8开始倒计时的评级，以调节场地距离主教的距离。(我希望这是英文的？(

然后你必须把矩形分成4个子。对于每个sub，您必须分别计算合法移动。

请尝试一下。如果你有问题，请告诉我。

public static void main(String[] args) throws Exception{
int pieces[][] = new int[8][8];
int res[][] = new int[8][8];
int x = 3;
int y = 4;
System.out.println(x + " XY " + y);
System.out.println("HAS RECT MOVES !");
System.out.println("_______ PIECES");
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
if(i<2){ pieces[i][j] = 1; }
if(i>5){ pieces[i][j] = 2; }
System.out.print(pieces[i][j] + " ");
}
System.out.println("");
}
System.out.println("_______ RES");
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
res[i][j] = checkField(i, j, x, y);
System.out.print(res[i][j] + " ");
}
System.out.println("");
}
System.out.println("My piece : " + (x) + " " + (y));
System.out.println("_______ legalmoves");
// 1. section:
System.out.println("1. section:");
calcSection(pieces, res, 0, 0, x, y);
// 2. section:
System.out.println("2. section:");
calcSection(pieces, res, 0, y, x, 7);
// 3. section:
System.out.println("3. section:");
calcSection(pieces, res, x, 0, 7, y);
// 4. section:
System.out.println("4. section:");
calcSection(pieces, res, x, y, 7, 7);
for (int i = 0; i < 8; i++) {
for (int j = 0; j < 8; j++) {
System.out.print(res[i][j] + " ");
}
System.out.println("");
}
}
public static int[][] calcSection(int[][] pieces, int[][] res, int recStartX, int recStartY, int recEndX, int recEndY){
System.out.println(recStartX+ ":"+ recStartY + ":"+ recEndX +":" +recEndY);
System.out.println("res[i][j] = " + res[5][7]);
for (int k = 8; k > 0; k--) {
for (int i = recStartX; i <= recEndX; i++) {
for (int j = recStartY; j <= recEndY; j++) {
if(res[j][i] == k){
if(pieces[j][i] > 0){
for (int l = recStartX; l <= recEndX; l++) {
for (int m = recStartY; m <= recEndY; m++) {
if(res[m][l] <= k){
res[m][l] = 0;
}
}
}       
return res;
}
}
}
}
}
return res;        
}
public static int checkField(int fieldX, int fieldY, int x, int y){
//x = 3;
//y = 2;
if((fieldY == x )&(fieldX == y )){
return 9;
}
else{
for (int i = 0; i < 8; i++) {
if((fieldY == x+i )&(fieldX == y+i )){
return 9-i;
} 
else if((fieldY == x+i )&(fieldX == y-i )){
return 9-i;
} 
else if((fieldY == x-i )&(fieldX == y+i )){
return 9-i;
}
else if((fieldY == x-i )&(fieldX == y-i )){
return 9-i;
}
}
}
return 0;
}
}