界面小游戏--五子棋

用到的知识

1.窗体界面

JFrame窗体的基本设置

2.事件机制

给窗体位置坐标加监听器，实现下棋的功能

3.图形的重绘，由于窗体显示，窗体会重绘两次，而画上去图像如果不加重绘方法，图像就不会显现，如最小化界面，移动界面等等，每移动一次，界面都是有系统调用重绘方法重新画界面，而上面的图像如果不定义重绘方法就不会再出现

4.数组你还没学过数组

? ? ? ? ? 神奇的数组，此小游戏中，用数组储存下过的旗子，黑白棋有标志分开，可以用于判断该交叉点是否有下过棋子，判断是否五子相连，数组用在此处很恰当，

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package chess;import java.awt.Color;import java.awt.Dimension;import java.awt.Graphics;import javax.swing.JFrame;public class ChessFrame extends JFrame{/** * @param args *///入口函数public static void main(String[] args) {// TODO Auto-generated method stubChessFrame frame=new ChessFrame();frame.initGUI();}//定义初始化界面方法public void initGUI(){this.setSize(new Dimension(580,600));this.setDefaultCloseOperation(3);this.setLocationRelativeTo(null);this.setVisible(true);//再创体显现之后，取得窗体画布图像Graphics g=this.getGraphics();//实例化处理器类ChessMouseListener cml=new ChessMouseListener(g);//给窗体加上监听器this.addMouseListener(cml);}//棋盘的重绘public void paint(Graphics g){//横线for(int i=0;i<Fig.LINE_H;i++){g.drawLine(Fig.X, Fig.Y+i*Fig.SIZE, Fig.X+Fig.SIZE*(Fig.LINE_H-1), Fig.Y+i*Fig.SIZE);}//竖线for(int i=0;i<Fig.LINE_V;i++){g.drawLine(Fig.X+i*Fig.SIZE, Fig.Y, Fig.X+i*Fig.SIZE, Fig.Y+Fig.SIZE*(Fig.LINE_V-1));}/****************旗子的重绘************************/for(int n=0;n<Fig.arr.length;n++){for(int m=0;m<Fig.arr[n].length;m++){//计算交叉点坐标int x=Fig.X+n*Fig.SIZE;int y=Fig.Y+m*Fig.SIZE;if(Fig.arr[n][m]==1){g.setColor(Color.BLACK);g.fillOval(x-Fig.SIZE/2, y-Fig.SIZE/2, Fig.SIZE, Fig.SIZE);}else if(Fig.arr[n][m]==-1){g.setColor(Color.WHITE);g.fillOval(x-Fig.SIZE/2, y-Fig.SIZE/2, Fig.SIZE, Fig.SIZE);}}}}}package chess;import java.awt.Color;import java.awt.Graphics;import java.awt.event.MouseAdapter;import java.awt.event.MouseEvent;public class ChessMouseListener extends MouseAdapter{//定义整形属性获取鼠标点击坐标private int x,y;//定义绘制棋子的属性private  Graphics g;//定义黑白棋的标志值private int state=1;/** *  * 构造方法，赋给属性的参数名 *///覆写父类方法public void mouseClicked(MouseEvent e){//取得点击的坐标值x=e.getX();y=e.getY();//循环进行下棋for(int i=0;i<Fig.LINE_H;i++){for(int j=0;j<Fig.LINE_V;j++){if(Fig.arr[i][j]==0){//判断交叉点的坐标int r=Fig.X+Fig.SIZE*i;int c=Fig.Y+Fig.SIZE*j;//计算出鼠标点击位置与交叉点的绝对值int tempX=Math.abs(r-x);int tempY=Math.abs(c-y);if(tempX<Fig.SIZE/2&&tempY<Fig.SIZE/2){x=r;y=c;//}else if(tempX<Fig.SIZE/2&&tempY>Fig.SIZE/2){//x=r;//y=c+Fig.SIZE;//}else if(tempX>Fig.SIZE/2&&tempY<Fig.SIZE/2){//x=r+Fig.SIZE;//y=c;//}else if(tempX>Fig.SIZE/2&&tempY>Fig.SIZE/2){//x=r+Fig.SIZE;//y=c+Fig.SIZE;//}//将旗子储存在数组中Fig.arr[i][j]=state;//判断是否下过棋子if(state==1){g.setColor(Color.BLACK);g.fillOval(x-Fig.SIZE/2, y-Fig.SIZE/2, Fig.SIZE, Fig.SIZE);state=-1;}else {g.setColor(Color.WHITE);g.fillOval(x-Fig.SIZE/2, y-Fig.SIZE/2, Fig.SIZE, Fig.SIZE);state=1;}Win win=new Win();boolean bool=win.validate(i, j);if(bool){if(state==1)System.out.println("白棋胜利");if(state==-1)System.out.println("黑棋胜利");}return;}}}}}public  ChessMouseListener(Graphics g){this.g=g;}}package chess;public interface Fig {//起始位置Xpublic static final int X=35;//起始位置Ypublic static final int Y=50;//横线public static final int LINE_H=15;//竖线public static final int LINE_V=15;//棋子大小public static final int CHESS=35;//网格大小public static final int SIZE=35;//存储棋子的数组public static final int arr[][]=new int[LINE_H][LINE_V];}package chess;public class Win {int count=1;public boolean validate(int r,int c){boolean state=false;//是否五子相连//if(winH(r,c)>=5){////如相连，返回真值//state=true;//}else{//state=false;//} //return state;//if(winH(r,c)>=5||winV(r,c)>=5||winR(r,c)>=5||winL(r,c)>=5){state=true;}return state;}//横向相连的个数public int winH(int r,int c){for(int i=r+1;i<Fig.arr.length;i++){if(Fig.arr[i][c]==Fig.arr[r][c]){count++;}else {break;}}for(int i1=r-1;i1>=0;i1--){if(Fig.arr[i1][c]==Fig.arr[r][c]){count++;}else {break;}}//System.out.println("棋相连个数"+count);return count;}//竖行相连的个数public int winV(int r,int c){//int count=1;for(int i=c+1;i<Fig.arr.length;i++){if(Fig.arr[r][i]==Fig.arr[r][c]){count++;}else {break;}}for(int i1=c-1;i1>=0;i1--){if(Fig.arr[r][i1]==Fig.arr[r][c]){count++;}else {break;}}return count;}//斜右public int winR(int r,int c){//int count=1;//int i,j;for(int i=r+1, j=c+1;i<Fig.arr.length&&j<Fig.arr[i].length;i++,j++){if(Fig.arr[i][j]==Fig.arr[r][c]){count++;}else {break;}}for(int i=r-1,j=c-1;i>=0&&j>=0;i--,j--){if(Fig.arr[i][j]==Fig.arr[r][c]){count++;}else {break;}}return count;}public int winL(int r,int c){//int count=1;int i,j;for( i=r+1, j=c-1;i<Fig.arr.length&&j>=0;i++,j--){if(Fig.arr[i][j]==Fig.arr[r][c]){count++;}else {break;}}for(i=r-1,j=c+1;i>=0&&j<Fig.arr[i].length;i--,j++){if(Fig.arr[i][j]==Fig.arr[r][c]){count++;}else {break;}}return count;}}

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