转载请注明出处:http://blog.csdn.net/einarzhang/article/details/44834259
本系列文章主要介绍如何利用Android开发一个自动生成题目的数独游戏。涉及的知识和技术如下所示:
- 挖洞算法自动生成数独题目
- 实现自定义View用于绘制数独盘
- 数据库的基本操作
看着市场上千篇一律的数独应用,他们大都来自于同一个开源应用,题目都是固定不变的那么100多道。我们就没有方法改变数独题目吗?经过百度搜索,终于找到了一篇自动生成数独题库的算法,感谢原作者的理论以及网络上的部分代码。算法文档题库:http://wenku.baidu.com/link?url=yA-IBTlo2hrjqcUOMcQ39ddYx1PaOnF0wjQycQTSYKIHJ5JUK-7WCK9Tz_BvDXQcio8I22k_xu1RZkwUYlUqFTZSa-jzyxgDfY3Ga93U34u
算法思想在文档中已经描述的十分具体,我们根据算法以及网络上的部分代码封装了可以生成任意难度数独的实体Sudoku,代码如下所示:
import java.util.ArrayList; import java.util.Random; /** * 采用挖洞算法实现。不同难度将会有不同的最小填充格子数和已知格子数 * */ public class Sudoku { private int[][] orginData; //保存初始状态的数独 // 初始化生成数组 private int[][] sudokuData = { { 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, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 0, 0, 0, 0, 0, 0, 0 } }; // 终盘数组 private int[][] resultData; // 最小填充格子数 private int minFilled; // 最小已知格子数 private int minKnow; // 当前难度级别;1-2简单难度;3普通难度;4高级难度;5骨灰级难度 private int level; private Random ran = new Random(); public Sudoku() { // this(3); // 默认为普通难度 } public Sudoku(int level) { if (level < 0 || level > 6) { this.level = 3; } else { this.level = level; } switch (level) { case 1: case 2: int ranNum = ran.nextInt(10); if(ranNum > 4) { minKnow = 5; } else { minKnow = 4; } minFilled = 45 + ranNum; break; // case 2: // minFilled = 45 + ran.nextInt(5); // minKnow = 4; // break; case 3: minFilled = 31 + ran.nextInt(10); minKnow = 3; break; case 4: minFilled = 21 + ran.nextInt(10); minKnow = 2; break; case 5: minFilled = 17 + ran.nextInt(10); minKnow = 0; break; default: break; } genSuduku(); orginData = new int[9][9]; for(int i = 0; i < 9; i++) { System.arraycopy(sudokuData[i], 0, orginData[i], 0, 9); } } /** * 生成唯一解的数独 */ public void genSuduku() { int startX = ran.nextInt(9), startY = ran.nextInt(9); // 初始挖洞格子 int orignStartX = startX, orignStartY = startY; // 暂存初始格子 int filledCount = 81; // 当前已知格子数 // int curMinKnow = 9; // 当前已知行列已知格子的下限 genShuduKnow(11); if (solve(sudokuData, true)) { // 将终盘赋值给初始数独数组 for (int i = 0; i < 9; i++) { System.arraycopy(resultData[i], 0, sudokuData[i], 0, 9); } // 实行等量交换 sudokuData = equalChange(sudokuData); Point nextP; // 下一个挖洞位置 do { int temMinKnow = getMinknow(sudokuData, startX, startY); if (isOnlyAnswer(sudokuData, startX, startY) && temMinKnow >= minKnow) { sudokuData[startX][startY] = 0; filledCount--; // curMinKnow = temMinKnow; } nextP = next(startX, startY); startX = nextP.x; startY = nextP.y; //校验此洞是否已挖 while(sudokuData[startX][startY] == 0 && (orignStartX != startX || orignStartY != startY)) { nextP = next(startX, startY); startX = nextP.x; startY = nextP.y; } //初级难度处理 if(level == 1 || level == 2) { while(startX == orignStartX && startY == orignStartY) { nextP = next(startX, startY); startX = nextP.x; startY = nextP.y; } } } while (filledCount > minFilled && (orignStartX != startX || orignStartY != startY)); } else { // 重新生成唯一解数独,直到成功为止 genSuduku(); } } /** * 获取下一个挖洞点 * * @param x * @param y * @return */ public Point next(int x, int y) { Point p = null; switch (level) { case 1: case 2: // 难度1、2均为随机算法 p = new Point(ran.nextInt(9), ran.nextInt(9)); break; case 3: // 难度3 间隔算法 if (x == 8 && y == 7) { p = new Point(0, 0); } else if (x == 8 && y == 8) { p = new Point(0, 1); } else if ((x % 2 == 0 && y == 7) || (x % 2 == 1) && y == 0) { p = new Point(x + 1, y + 1); } else if ((x % 2 == 0 && y == 8) || (x % 2 == 1) && y == 1) { p = new Point(x + 1, y - 1); } else if (x % 2 == 0) { p = new Point(x, y + 2); } else if (x % 2 == 1) { p = new Point(x, y - 2); } break; case 4: // 难度4 蛇形算法 p = new Point(); if (x == 8 && y == 8) { p.y = 0; } else if (x % 2 == 0 && y < 8) { // 蛇形顺序,对下个位置列的求解 p.y = y + 1; } else if ((x % 2 == 0 && y == 8) || (x % 2 == 1 && y == 0)) { p.y = y; } else if (x % 2 == 1 && y > 0) { p.y = y - 1; } if (x == 8 && y == 8) { // 蛇形顺序,对下个位置行的求解 p.x = 0; } else if ((x % 2 == 0 && y == 8) || (x % 2 == 1) && y == 0) { p.x = x + 1; } else { p.x = x; } break; case 5: // 从左至右,从上到下 p = new Point(); if (y == 8) { if (x == 8) { p.x = 0; } else { p.x = x + 1; } p.y = 0; } else { p.x = x; p.y = y + 1; } default: break; } return p; } /** * 生成指定个数的已知格子 * * @param n */ public void genShuduKnow(int n) { // 生成n个已知格子 Point[] knowPonits = new Point[n]; for (int i = 0; i < n; i++) { knowPonits[i] = new Point(ran.nextInt(9), ran.nextInt(9)); // 检测是否重复 for (int k = 0; k < i; k++) { if (knowPonits[k].equals(knowPonits[i])) { i--; break; } } } // 填充数字 int num; Point p; for (int i = 0; i < n; i++) { num = 1 + ran.nextInt(9); p = knowPonits[i]; sudokuData[p.x][p.y] = num; if (!validateIandJ(sudokuData, p.x, p.y)) { // 生成格子填充数字错误 i--; } } } /** * 等量交换 * * @param data * @return */ public int[][] equalChange(int[][] data) { Random rand = new Random(); int num1 = 1 + rand.nextInt(9); int num2 = 1 + rand.nextInt(9); for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { if (data[i][j] == 1) { data[i][j] = num1; } else if (data[i][j] == num1) { data[i][j] = 1; } if (data[i][j] == 2) { data[i][j] = num2; } else if (data[i][j] == num2) { data[i][j] = 2; } } } return data; } /** * 判断挖去i,j位置后,是否有唯一解 * * @param data * @param i * @param j * @return */ public boolean isOnlyAnswer(int[][] data, int i, int j) { int k = data[i][j]; // 待挖洞的原始数字 for (int num = 1; num < 10; num++) { data[i][j] = num; if (num != k && solve(data, false)) { // 除待挖的数字之外,还有其他的解,则返回false data[i][j] = k; return false; } } data[i][j] = k; return true; } /** * 删除指定位置后,新的行列下限 * * @param data * @param p * @param q * @return */ public int getMinknow(int[][] data, int p, int q) { int temp = data[p][q]; int minKnow = 9; int tempKnow = 9; data[p][q] = 0; for (int i = 0; i < 9; i++) { // 行下限 for (int j = 0; j < 9; j++) { if (data[i][j] == 0) { tempKnow--; if (tempKnow < minKnow) { minKnow = tempKnow; } } } tempKnow = 9; } tempKnow = 9; for (int j = 0; j < 9; j++) { // 列下限 for (int i = 0; i < 9; i++) { if (data[i][j] == 0) { tempKnow--; if (tempKnow < minKnow) { minKnow = tempKnow; } } } tempKnow = 9; } data[p][q] = temp; // 返回删除后的下限 return minKnow; } /** * 判断数独是否能解 * * @param data * @return */ public boolean solve(int[][] data, boolean flag) { int blankCount = 0; // 保存空位个数 int[][] tempData = new int[9][9]; // 中间数组 ArrayList<Point> blankList = new ArrayList<Point>(70); // 保存各个空格坐标 for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { tempData[i][j] = data[i][j]; if (tempData[i][j] == 0) { blankCount++; blankList.add(new Point(i, j)); } } } // 校验数独合法性 if (!validate(tempData)) { return false; } if (blankCount == 0) { // 玩家已经成功解出数独 return true; } else if (put(tempData, 0, blankList)) { if(flag) { // 智能解出答案,供生成数独终盘使用 resultData = tempData; } return true; } return false; } /** * 在第n个空位子放入数字 * * @param data * @param n * @param blankList * @return */ public boolean put(int[][] data, int n, ArrayList<Point> blankList) { if (n < blankList.size()) { Point p = blankList.get(n); for (int i = 1; i < 10; i++) { data[p.x][p.y] = i; if (validateIandJ(data, p.x, p.y) && put(data, n + 1, blankList)) { return true; } } data[p.x][p.y] = 0; return false; } else { return true; } } /** * 校验x, y位置填充的数字是否可行 * * @param data * @param x * @param y * @return */ public boolean validateIandJ(int[][] data, int x, int y) { int m = 0, n = 0, p = 0, q = 0; // m,n是计数器,p,q用于确定测试点的方格位置 for (m = 0; m < 9; m++) { if (m != x && data[m][y] == data[x][y]) { return false; } } for (n = 0; n < 9; n++) { if (n != y && data[x][n] == data[x][y]) { return false; } } for (p = x / 3 * 3, m = 0; m < 3; m++) { for (q = y / 3 * 3, n = 0; n < 3; n++) { if ((p + m != x || q + n != y) && (data[p + m][q + n] == data[x][y])) { return false; } } } return true; } /** * 校验整个数独数组的合法性 * * @param data * @return */ public boolean validate(int[][] data) { for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { if (data[i][j] != 0 && !validateIandJ(data, i, j)) { // 任何一个数字填充错误,则返回false return false; } } } return true; } /** * 获取冲突列表 * @return */ public ArrayList<Point> validateForList() { ArrayList<Point> pList = new ArrayList<Point>(); for (int i = 0; i < 9; i++) { for (int j = 0; j < 9; j++) { if (sudokuData[i][j] != 0 && !validateIandJ(sudokuData, i, j)) { pList.add(new Point(i, j)); } } } return pList; } public int[][] myResultData() { return resultData; } public int[][] myInitData() { return orginData; } public int[][] mySudoku() { return sudokuData; } /** * 判断格子是否为已知格子 * @param x * @param y * @return */ public boolean isKnownCell(int x, int y) { return orginData[x][y] != 0; } public void makeToInitData() { for(int i = 0; i < 9; i++) { System.arraycopy(orginData[i], 0, sudokuData[i], 0, 9); } } public void set(int x, int y, int num) { sudokuData[x][y] = num; } public boolean isSuccess() { for(int i = 0; i < 9; i++) { for(int j = 0; j < 9; j++) { // if(sudokuData[i][j] == 0 || sudokuData[i][j] != resultData[i][j]) { // return false; // } if(sudokuData[i][j] == 0) { return false; } } } return validate(sudokuData); // return true; } public void setOrginData(int[][] orginData) { this.orginData = orginData; } public void setSudokuData(int[][] sudokuData) { this.sudokuData = sudokuData; } public void setResultData(int[][] resultData) { this.resultData = resultData; } }
时间: 2024-10-18 04:48:28