求java小游戲源代碼

表1. CheckerDrag.java

// CheckerDrag.javaimport java.awt.*;import java.awt.event.*;public class CheckerDrag extends java.applet.Applet{ // Dimension of checkerboard square. // 棋盤上每個(gè)小方格的尺寸 final static int SQUAREDIM = 40; // Dimension of checkerboard -- includes black outline. // 棋盤的尺寸 – 包括黑色的輪廓線 final static int BOARDDIM = 8 * SQUAREDIM + 2; // Dimension of checker -- 3/4 the dimension of a square. // 棋子的尺寸 – 方格尺寸的3/4 final static int CHECKERDIM = 3 * SQUAREDIM / 4; // Square colors are dark green or white. // 方格的顏色為深綠色或者白色 final static Color darkGreen = new Color (0, 128, 0); // Dragging flag -- set to true when user presses mouse button over checker // and cleared to false when user releases mouse button. // 拖動(dòng)標(biāo)記 --當(dāng)用戶在棋子上按下鼠標(biāo)按鍵時(shí)設(shè)為true， // 釋放鼠標(biāo)按鍵時(shí)設(shè)為false boolean inDrag = false; // Left coordinate of checkerboard's upper-left corner. // 棋盤左上角的左方向坐標(biāo) int boardx; // Top coordinate of checkerboard's upper-left corner. //棋盤左上角的上方向坐標(biāo) int boardy; // Left coordinate of checker rectangle origin (upper-left corner). // 棋子矩形原點(diǎn)(左上角)的左方向坐標(biāo) int ox; // Top coordinate of checker rectangle origin (upper-left corner). // 棋子矩形原點(diǎn)(左上角)的上方向坐標(biāo) int oy; // Left displacement between mouse coordinates at time of press and checker // rectangle origin. // 在按鍵時(shí)的鼠標(biāo)坐標(biāo)與棋子矩形原點(diǎn)之間的左方向位移 int relx; // Top displacement between mouse coordinates at time of press and checker // rectangle origin. // 在按鍵時(shí)的鼠標(biāo)坐標(biāo)與棋子矩形原點(diǎn)之間的上方向位移 int rely; // Width of applet drawing area. // applet繪圖區(qū)域的寬度 int width; // Height of applet drawing area. // applet繪圖區(qū)域的高度 int height; // Image buffer. // 圖像緩沖 Image imBuffer; // Graphics context associated with image buffer. // 圖像緩沖相關(guān)聯(lián)的圖形背景 Graphics imG; public void init () { // Obtain the size of the applet's drawing area. // 獲取applet繪圖區(qū)域的尺寸 width = getSize ().width; height = getSize ().height; // Create image buffer. // 創(chuàng)建圖像緩沖 imBuffer = createImage (width, height); // Retrieve graphics context associated with image buffer. // 取出圖像緩沖相關(guān)聯(lián)的圖形背景 imG = imBuffer.getGraphics (); // Initialize checkerboard's origin, so that board is centered. // 初始化棋盤的原點(diǎn)，使棋盤在屏幕上居中 boardx = (width - BOARDDIM) / 2 + 1; boardy = (height - BOARDDIM) / 2 + 1; // Initialize checker's rectangle's starting origin so that checker is // centered in the square located in the top row and second column from // the left. // 初始化棋子矩形的起始原點(diǎn)，使得棋子在第一行左數(shù)第二列的方格里居中 ox = boardx + SQUAREDIM + (SQUAREDIM - CHECKERDIM) / 2 + 1; oy = boardy + (SQUAREDIM - CHECKERDIM) / 2 + 1; // Attach a mouse listener to the applet. That listener listens for // mouse-button press and mouse-button release events. // 向applet添加一個(gè)用來(lái)監(jiān)聽鼠標(biāo)按鍵的按下和釋放事件的鼠標(biāo)監(jiān)聽器 addMouseListener (new MouseAdapter () { public void mousePressed (MouseEvent e) { // Obtain mouse coordinates at time of press. // 獲取按鍵時(shí)的鼠標(biāo)坐標(biāo) int x = e.getX (); int y = e.getY (); // If mouse is over draggable checker at time // of press (i.e., contains (x, y) returns // true), save distance between current mouse // coordinates and draggable checker origin // (which will always be positive) and set drag // flag to true (to indicate drag in progress). // 在按鍵時(shí)如果鼠標(biāo)位于可拖動(dòng)的棋子上方 // （也就是contains (x, y)返回true），則保存當(dāng)前 // 鼠標(biāo)坐標(biāo)與棋子的原點(diǎn)之間的距離（始終為正值）并且 // 將拖動(dòng)標(biāo)志設(shè)為true（用來(lái)表明正處在拖動(dòng)過程中） if (contains (x, y)) { relx = x - ox; rely = y - oy; inDrag = true; } } boolean contains (int x, int y) { // Calculate center of draggable checker. // 計(jì)算棋子的中心位置 int cox = ox + CHECKERDIM / 2; int coy = oy + CHECKERDIM / 2; // Return true if (x, y) locates with bounds // of draggable checker. CHECKERDIM / 2 is the // radius. // 如果(x, y)仍處于棋子范圍內(nèi)則返回true // CHECKERDIM / 2為半徑 return (cox - x) * (cox - x) + (coy - y) * (coy - y) CHECKERDIM / 2 * CHECKERDIM / 2; } public void mouseReleased (MouseEvent e) { // When mouse is released, clear inDrag (to // indicate no drag in progress) if inDrag is // already set. // 當(dāng)鼠標(biāo)按鍵被釋放時(shí)，如果inDrag已經(jīng)為true， // 則將其置為false（用來(lái)表明不在拖動(dòng)過程中） if (inDrag) inDrag = false; } }); // Attach a mouse motion listener to the applet. That listener listens // for mouse drag events. //向applet添加一個(gè)用來(lái)監(jiān)聽鼠標(biāo)拖動(dòng)事件的鼠標(biāo)運(yùn)動(dòng)監(jiān)聽器 addMouseMotionListener (new MouseMotionAdapter () { public void mouseDragged (MouseEvent e) { if (inDrag) { // Calculate draggable checker's new // origin (the upper-left corner of // the checker rectangle). // 計(jì)算棋子新的原點(diǎn)（棋子矩形的左上角） int tmpox = e.getX () - relx; int tmpoy = e.getY () - rely; // If the checker is not being moved // (at least partly) off board, // assign the previously calculated // origin (tmpox, tmpoy) as the // permanent origin (ox, oy), and // redraw the display area (with the // draggable checker at the new // coordinates). // 如果棋子（至少是棋子的一部分）沒有被 // 移出棋盤，則將之前計(jì)算的原點(diǎn) // (tmpox, tmpoy)賦值給永久性的原點(diǎn)(ox, oy)， // 并且刷新顯示區(qū)域（此時(shí)的棋子已經(jīng)位于新坐標(biāo)上） if (tmpox boardx tmpoy boardy tmpox + CHECKERDIM boardx + BOARDDIM tmpoy + CHECKERDIM boardy + BOARDDIM) { ox = tmpox; oy = tmpoy; repaint (); } } } }); } public void paint (Graphics g) { // Paint the checkerboard over which the checker will be dragged. // 在棋子將要被拖動(dòng)的位置上繪制棋盤 paintCheckerBoard (imG, boardx, boardy); // Paint the checker that will be dragged. // 繪制即將被拖動(dòng)的棋子 paintChecker (imG, ox, oy); // Draw contents of image buffer. // 繪制圖像緩沖的內(nèi)容 g.drawImage (imBuffer, 0, 0, this); } void paintChecker (Graphics g, int x, int y) { // Set checker shadow color. // 設(shè)置棋子陰影的顏色 g.setColor (Color.black); // Paint checker shadow. // 繪制棋子的陰影 g.fillOval (x, y, CHECKERDIM, CHECKERDIM); // Set checker color. // 設(shè)置棋子顏色 g.setColor (Color.red); // Paint checker. // 繪制棋子 g.fillOval (x, y, CHECKERDIM - CHECKERDIM / 13, CHECKERDIM - CHECKERDIM / 13); } void paintCheckerBoard (Graphics g, int x, int y) { // Paint checkerboard outline. // 繪制棋盤輪廓線 g.setColor (Color.black); g.drawRect (x, y, 8 * SQUAREDIM + 1, 8 * SQUAREDIM + 1); // Paint checkerboard. // 繪制棋盤 for (int row = 0; row 8; row++) { g.setColor (((row 1) != 0) ? darkGreen : Color.white); for (int col = 0; col 8; col++) { g.fillRect (x + 1 + col * SQUAREDIM, y + 1 + row * SQUAREDIM, SQUAREDIM, SQUAREDIM); g.setColor ((g.getColor () == darkGreen) ? Color.white : darkGreen); } } } // The AWT invokes the update() method in response to the repaint() method // calls that are made as a checker is dragged. The default implementation // of this method, which is inherited from the Container class, clears the // applet's drawing area to the background color prior to calling paint(). // This clearing followed by drawing causes flicker. CheckerDrag overrides // update() to prevent the background from being cleared, which eliminates // the flicker. // AWT調(diào)用了update()方法來(lái)響應(yīng)拖動(dòng)棋子時(shí)所調(diào)用的repaint()方法。該方法從 // Container類繼承的默認(rèn)實(shí)現(xiàn)會(huì)在調(diào)用paint()之前，將applet的繪圖區(qū)域清除 // 為背景色，這種繪制之后的清除就導(dǎo)致了閃爍。CheckerDrag重寫了update()來(lái) // 防止背景被清除，從而消除了閃爍。 public void update (Graphics g) { paint (g); }}

用java語(yǔ)言編寫連連看游戲

我以前自己寫一個(gè)玩的。沒有把你要求的功能全部實(shí)現(xiàn)，不過你看了后可以改一下就好了。參考一下吧，我給了注解：

package mybase.programe;

/*

* lianliankan總體算法思路：由兩個(gè)確定的按鈕。若這兩個(gè)按鈕的數(shù)字相等，就開始找它們相連的路經(jīng)。這個(gè)找路經(jīng)

* 分3種情況：(從下面的這三種情況，我們可以知道，需要三個(gè)檢測(cè)，這三個(gè)檢測(cè)分別檢測(cè)一條直路經(jīng)。這樣就會(huì)有

* 三條路經(jīng)。若這三條路經(jīng)上都是空按鈕，那么就剛好是三種直線（兩個(gè)轉(zhuǎn)彎點(diǎn)）把兩個(gè)按鈕連接起來(lái)了)

* 1.相鄰

*

* 2. 若不相鄰的先在第一個(gè)按鈕的同行找一個(gè)空按鈕。1).找到后看第二個(gè)按鈕橫向到這個(gè)空按鈕

* 所在的列是否有按鈕。2).沒有的話再看第一個(gè)按鈕到與它同行的那個(gè)空按鈕之間是否有按鈕。3).沒有的話，再?gòu)?/p>

* 與第一個(gè)按鈕同行的那個(gè)空按鈕豎向到與第二個(gè)按鈕的同行看是否有按鈕。沒有的話路經(jīng)就通了,可以消了.

*

* 3.若2失敗后，再在第一個(gè)按鈕的同列找一個(gè)空按鈕。1).找到后看第二個(gè)按鈕豎向到這個(gè)空按鈕所在的行是否有按鈕。

* 2).沒有的話，再看第一個(gè)按鈕到與它同列的那個(gè)空按鈕之間是否有按鈕。3).沒有的話，再?gòu)呐c第一個(gè)按鈕同列的

* 那個(gè)空按鈕橫向到與第二個(gè)按鈕同列看是否有按鈕。沒有的話路經(jīng)就通了，可以消了。

*

* 若以上三步都失敗，說明這兩個(gè)按鈕不可以消去。

*/

import javax.swing.*;

import java.awt.*;

import java.awt.event.*;

public class LianLianKan implements ActionListener {

JFrame mainFrame; // 主面板

Container thisContainer;

JPanel centerPanel, southPanel, northPanel; // 子面板

JButton diamondsButton[][] = new JButton[6][5];// 游戲按鈕數(shù)組

JButton exitButton, resetButton, newlyButton; // 退出，重列，重新開始按鈕

JLabel fractionLable = new JLabel("0"); // 分?jǐn)?shù)標(biāo)簽

JButton firstButton, secondButton; // 分別記錄兩次被選中的按鈕

// 儲(chǔ)存游戲按鈕位置(這里其實(shí)只要6行，5列。但是我們用了8行，7列。是等于在這個(gè)面板按鈕的周圍還圍

//了一層是0的按鈕,這樣就可以實(shí)現(xiàn)靠近面板邊緣的兩個(gè)按鈕可以消去)

int grid[][] = new int[8][7];

static boolean pressInformation = false; // 判斷是否有按鈕被選中

int x0 = 0, y0 = 0, x = 0, y = 0, fristMsg = 0, secondMsg = 0, validateLV; // 游戲按鈕的位置坐標(biāo)

int i, j, k, n;// 消除方法控制

public void init() {

mainFrame = new JFrame("JKJ連連看");

thisContainer = mainFrame.getContentPane();

thisContainer.setLayout(new BorderLayout());

centerPanel = new JPanel();

southPanel = new JPanel();

northPanel = new JPanel();

thisContainer.add(centerPanel, "Center");

thisContainer.add(southPanel, "South");

thisContainer.add(northPanel, "North");

centerPanel.setLayout(new GridLayout(6, 5));

for (int cols = 0; cols 6; cols++) {

for (int rows = 0; rows 5; rows++) {

diamondsButton[cols][rows] = new JButton(String

.valueOf(grid[cols + 1][rows + 1]));

diamondsButton[cols][rows].addActionListener(this);

centerPanel.add(diamondsButton[cols][rows]);

}

}

exitButton = new JButton("退出");

exitButton.addActionListener(this);

resetButton = new JButton("重列");

resetButton.addActionListener(this);

newlyButton = new JButton("再來(lái)一局");

newlyButton.addActionListener(this);

southPanel.add(exitButton);

southPanel.add(resetButton);

southPanel.add(newlyButton);

fractionLable.setText(String.valueOf(Integer.parseInt(fractionLable

.getText())));

northPanel.add(fractionLable);

mainFrame.setBounds(280, 100, 500, 450);

mainFrame.setVisible(true);

mainFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

}

public void randomBuild() {

int randoms, cols, rows;

for (int twins = 1; twins = 15; twins++) {//一共15對(duì)button,30個(gè)

randoms = (int) (Math.random() * 25 + 1);//button上的數(shù)字

for (int alike = 1; alike = 2; alike++) {

cols = (int) (Math.random() * 6 + 1);

rows = (int) (Math.random() * 5 + 1);

while (grid[cols][rows] != 0) {//等于0說明這個(gè)空格有了button

cols = (int) (Math.random() * 6 + 1);

rows = (int) (Math.random() * 5 + 1);

}

this.grid[cols][rows] = randoms;

}

}

}

public void fraction() {

fractionLable.setText(String.valueOf(Integer.parseInt(fractionLable

.getText()) + 100));

}

public void reload() {

int save[] = new int[30];

int n = 0, cols, rows;

int grid[][] = new int[8][7];

for (int i = 0; i = 6; i++) {

for (int j = 0; j = 5; j++) {

if (this.grid[i][j] != 0) {

save[n] = this.grid[i][j];//記下每個(gè)button的數(shù)字

n++;//有幾個(gè)沒有消去的button

}

}

}

n = n - 1;

this.grid = grid;

while (n = 0) {//把沒有消去的button重新放一次

cols = (int) (Math.random() * 6 + 1);

rows = (int) (Math.random() * 5 + 1);

while (grid[cols][rows] != 0) {

cols = (int) (Math.random() * 6 + 1);

rows = (int) (Math.random() * 5 + 1);

}

this.grid[cols][rows] = save[n];

n--;

}

mainFrame.setVisible(false);

pressInformation = false; // 這里一定要將按鈕點(diǎn)擊信息歸為初始

init();

for (int i = 0; i 6; i++) {

for (int j = 0; j 5; j++) {

if (grid[i + 1][j + 1] == 0)

diamondsButton[i][j].setVisible(false);

}

}

}

public void estimateEven(int placeX, int placeY, JButton bz) {

if (pressInformation == false) {

x = placeX;

y = placeY;

secondMsg = grid[x][y];

secondButton = bz;

pressInformation = true;

} else {

x0 = x;

y0 = y;

fristMsg = secondMsg;

firstButton = secondButton;

x = placeX;

y = placeY;

secondMsg = grid[x][y];

secondButton = bz;

if (fristMsg == secondMsg secondButton != firstButton) {

xiao();

}

}

}

public void xiao() { // 相同的情況下能不能消去。仔細(xì)分析，不一條條注釋

if ((x0 == x (y0 == y + 1 || y0 == y - 1))

|| ((x0 == x + 1 || x0 == x - 1) (y0 == y))) { // 判斷是否相鄰

remove();

} else {

for (j = 0; j 7; j++) {

if (grid[x0][j] == 0) { // 判斷和第一個(gè)按鈕同行的哪個(gè)按鈕為空

//如果找到一個(gè)為空的，就按列值的三種情況比較第二個(gè)按鈕與空按鈕的位置

if (y j) {//第二個(gè)按鈕在空按鈕右邊

for (i = y - 1; i = j; i--) { //檢測(cè)從第二個(gè)按鈕橫向左邊到空格所在列為止是否全是空格

if (grid[x][i] != 0) {

k = 0;

break;//存在非空格的就退出，這一退出就不可能k==2了，所以就會(huì)到下而215行出同理的判斷列

} else {

k = 1;

} // K=1說明全是空格通過了第一次驗(yàn)證，也就是從第二個(gè)按鈕橫向左邊到空格所在列為止全是空格

}

if (k == 1) {

linePassOne();//進(jìn)入第二次驗(yàn)證，也就是從第一個(gè)按鈕到它同行的空格之間的空格判斷

}

}

if (y j) { // 第二個(gè)按鈕在空按鈕左邊

for (i = y + 1; i = j; i++) {//檢測(cè)從第二個(gè)按鈕橫向右邊到空格所在列為止是否全是空格

if (grid[x][i] != 0) {

k = 0;

break;

} else {

k = 1;

}

}

if (k == 1) {

linePassOne();

}

}

if (y == j) {//第二個(gè)按鈕和空按鈕同列

linePassOne();

}

}

//第三次檢測(cè)，檢測(cè)確定為空的第j列的那個(gè)按鈕豎向到第二個(gè)按鈕，看是不是有按鈕

if (k == 2) {

if (x0 == x) {//第一，二按鈕在同行

remove();

}

if (x0 x) {//第一按鈕在第二按鈕下邊

for (n = x0; n = x - 1; n++) {//從空按鈕豎向到第二個(gè)按鈕所在行是否有按鈕

if (grid[n][j] != 0) {

k = 0;

break;

}

//沒有按鈕，說明這條路經(jīng)就通了

if (grid[n][j] == 0 n == x - 1) {

remove();

}

}

}

if (x0 x) {//第一按鈕在第二按鈕上邊

for (n = x0; n = x + 1; n--) {

if (grid[n][j] != 0) {

k = 0;

break;

}

if (grid[n][j] == 0 n == x + 1) {

remove();

}

}

}

}

}//-------------------------------------for

//當(dāng)上面的檢測(cè)與第一個(gè)按鈕同行的空格按鈕失敗后(不能找到與第二個(gè)按鈕的相連路經(jīng))，下面就執(zhí)行

//檢測(cè)與第一個(gè)按鈕同列的空格按鈕

for (i = 0; i 8; i++) {

if (grid[i][y0] == 0) {// 判斷和第一個(gè)按鈕同列的哪個(gè)按鈕為空

if (x i) {//第二個(gè)按鈕在這個(gè)空按鈕的下面

for (j = x - 1; j = i; j--) {

if (grid[j][y] != 0) {

k = 0;

break;

} else {

k = 1;

}

}

if (k == 1) {

rowPassOne();

}

}

if (x i) {//第二個(gè)按鈕在這個(gè)空按鈕的上面

for (j = x + 1; j = i; j++) {

if (grid[j][y] != 0) {

k = 0;

break;

} else {

k = 1;

}

}

if (k == 1) {

rowPassOne();

}

}

if (x == i) {//第二個(gè)按鈕與這個(gè)空按鈕同行

rowPassOne();

}

}

if (k == 2) {

if (y0 == y) {//第二個(gè)按鈕與第一個(gè)按鈕同列

remove();

}

if (y0 y) {//第二個(gè)按鈕在第一個(gè)按鈕右邊

for (n = y0; n = y - 1; n++) {

if (grid[i][n] != 0) {

k = 0;

break;

}

if (grid[i][n] == 0 n == y - 1) {

remove();

}

}

}

if (y0 y) {//第二個(gè)按鈕在第一個(gè)按鈕左邊

for (n = y0; n = y + 1; n--) {

if (grid[i][n] != 0) {

k = 0;

break;

}

if (grid[i][n] == 0 n == y + 1) {

remove();

}

}

}

}

}//--------------------------------for

}//-------------else

}//------------xiao

public void linePassOne() {

if (y0 j) { // 第一按鈕同行空按鈕在左邊

for (i = y0 - 1; i = j; i--) { // 判斷第一按鈕同左側(cè)空按鈕之間有沒按鈕

if (grid[x0][i] != 0) {

k = 0;

break;

} else {

k = 2;

} // K=2說明通過了第二次驗(yàn)證

}

}

if (y0 j) { // 第一按鈕同行空按鈕在右邊

for (i = y0 + 1; i = j; i++) {

if (grid[x0][i] != 0) {

k = 0;

break;

} else {

k = 2;

}

}

}

}

public void rowPassOne() {

if (x0 i) {//第一個(gè)按鈕在與它同列的那個(gè)空格按鈕下面

for (j = x0 - 1; j = i; j--) {

if (grid[j][y0] != 0) {

k = 0;

break;

} else {

k = 2;

}

}

}

if (x0 i) {//第一個(gè)按鈕在與它同列的那個(gè)空格按鈕上面

for (j = x0 + 1; j = i; j++) {

if (grid[j][y0] != 0) {

k = 0;

break;

} else {

k = 2;

}

}

}

}

public void remove() {

firstButton.setVisible(false);

secondButton.setVisible(false);

fraction();

pressInformation = false;

k = 0;

grid[x0][y0] = 0;

grid[x][y] = 0;

}

public void actionPerformed(ActionEvent e) {

if (e.getSource() == newlyButton) {

int grid[][] = new int[8][7];

this.grid = grid;

randomBuild();

mainFrame.setVisible(false);

pressInformation = false;

init();

}

if (e.getSource() == exitButton)

System.exit(0);

if (e.getSource() == resetButton)

reload();

for (int cols = 0; cols 6; cols++) {

for (int rows = 0; rows 5; rows++) {

if (e.getSource() == diamondsButton[cols][rows])

estimateEven(cols + 1, rows + 1, diamondsButton[cols][rows]);

}

}

}

public static void main(String[] args) {

LianLianKan llk = new LianLianKan();

llk.randomBuild();

llk.init();

}

}

連連看JAVA源代碼

加上。(初始化代碼樓主清洗本身選) 對(duì)應(yīng)在這句話。別離grid[][]數(shù)組的行列即可，你只需定義25個(gè)不一樣的圖片;后面; 這句話是用來(lái)設(shè)置連連看的圖的.setIcon(icons[grid[cols + 1][rows + 1])])。 定義： diamondsButton[cols][rows] = new JButton(String .valueOf(grid[cols + 1][rows + 1]))： ImageIcon icons[]= new ImageIcon[25]： diamondsButton[cols][rows].valueOf(grid[cols + 1][rows + 1]))diamondsButton[cols][rows] = new JButton(String 。它只用了數(shù)字; 然后把icons數(shù)組初始化對(duì)應(yīng)每個(gè)圖片即可