连连看算法
第一步:我们考虑在同行或者同列的情况:
同行或者同列又分三种情况:
第一:边线,RowNum或者ColNum等于0或者9
第二:两个相邻
第三:同行不相邻,这种事有条件的,所在行(或列)的上下(或左右)紧邻行必须在两个按钮之间已全部消除?好,有出问题了,怎么根据已知的行和列,获取Button是否被消除呢?我们可以
定义一个Button的二维数组。并在初始化的时候给它赋值
MyButton[,] maps=new MyButton[10,10];
maps[i, j] = btn;
好,怎么获取Button是否隐藏已经解决。
同行步相邻的又分是从上侧、右侧和中间连。
第二步:既不同行又不同列的要考虑的更多,东西南北都要考虑到了,而连连看的规则是最多转两个弯,
好了不说了直接上代码了:算法可以自己慢慢琢磨(如果想要最优算法你可以考虑研究下最短路径算法)
private bool Check(MyButton btn1, MyButton btn2) { for (int c = 0; c < 10; c++) { if (c != btn1.ColNum && !(btn1.RowNum == btn2.RowNum && btn1.ColNum == c) && maps[btn1.RowNum, c].Visibility == Visibility.Visible) //不为A和B的障碍 continue; if (!IsLine(btn1.RowNum, btn2.RowNum, btn1.RowNum, c)) //Aa continue; for (int c2 = 0; c2 < 10; c2++) { if (c2 != btn2.ColNum && !(btn1.RowNum == btn2.RowNum && btn1.ColNum == c2) && maps[btn2.RowNum, c2].Visibility == Visibility.Visible) continue; if (!IsLine(btn1.RowNum, c, btn1.RowNum, c2)) //ab continue; if (!IsLine(btn2.RowNum, c2, btn2.RowNum, btn2.ColNum)) //bB continue; return true; } for (int r = 0; r < 10; r++) { if (r != btn2.RowNum && !(r == btn1.RowNum && btn2.ColNum == btn1.ColNum) && maps[r, btn2.ColNum].Visibility == Visibility.Visible) continue; if (!IsLine(btn1.RowNum, c, r, btn2.ColNum)) //ab continue; if (!IsLine(r, btn2.ColNum, btn2.RowNum, btn2.ColNum)) //bB continue; return true; } } for (int r = 0; r < 10; r++) { if (r != btn1.RowNum && !(r == btn2.RowNum && btn1.ColNum == btn2.ColNum) && maps[r, btn1.RowNum].Visibility == Visibility.Visible) //不为A和B的障碍 continue; if (!IsLine(btn1.RowNum, btn1.ColNum, r, btn1.ColNum)) //Aa continue; for (int c2 = 0; c2 < 10; c2++) { if (c2 != btn2.ColNum && maps[btn2.RowNum, c2].Visibility == Visibility.Visible) continue; if (!IsLine(r, btn1.RowNum, btn2.RowNum, c2)) //ab continue; if (!IsLine(btn2.RowNum, c2, btn2.RowNum, btn2.ColNum)) //bB continue; return true; } for (int r2 = 0; r2 < 10; r2++) { if (r2 != btn2.RowNum && !(r2 == btn1.RowNum && btn2.ColNum == btn1.ColNum) && maps[r2, btn2.ColNum].Visibility == Visibility.Visible) continue; if (!IsLine(r, btn1.ColNum, r2, btn2.ColNum)) //ab continue; if (!IsLine(r2, btn2.ColNum, btn2.RowNum, btn2.ColNum)) //bB continue; return true; } } return false; } private bool IsLine(int r1, int c1, int r2, int c2) { int max, min; if (r1 == r2) //同行 { max = ((c1 > c2) ? c1 : c2) - 1; min = ((c1 > c2) ? c2 : c1) + 1; while (min <= max) { if (maps[r1, min].Visibility == Visibility.Visible) return false; min++; } return true; } if (c1 == c2) //同列 { max = ((r1 > r2) ? r1 : r2) - 1; min = ((r1 > r2) ? r2 : r1) + 1; while (min <= max) { if (maps[min, c1].Visibility == Visibility.Visible) return false; min++; } return true; } return false; }
第三步:上面我们完成了算法,也就是我们基本可以玩了。但是玩的过程我们又发现了新问题,
我们没有保证每个图像都是偶数个,如图:
这个问题要从初始化的时候着手,我们现在要做一百个背景,那么我们先做50个,然后再分别从这个50个中取出
代码如下:
public List<int> Get() { Random rand = new Random(); List<int> ran = new List<int>(); for (int n = 0; n < 50; n++) ran.Add(rand.Next(1, 10)); for (int i = 1; i < 50; i++) { List<int> temp = new List<int>(); temp.AddRange(ran); for (int n = 0; n < 50; n++) { int r = rand.Next(0, temp.Count); ran.Insert(rand.Next(0, ran.Count), temp[r]); temp.RemoveAt(r); } } return ran; }
这种思路应该是对的,但是没有达到预期的效果,还是会有剩下,由于缺少时间,可能要过一段时间来弄这个,有
看出问题的高手可以帮我留言,我会争取早点解决问题的。
我下一步准备是将动画连线与倒计时加上,这可能要等一段时间了。下面是我在网上找的一段练练看算法,希望
能给大家一个参考:
public class LineCore { //整体宽度 public int width { get; set; } //整体高度 public int height { get; set; } //前次被点击的图片 public Position lastPicture { get; set; } //当前被电击的图片 public Position currentPicture { get; set; } // 图片矩阵,-1表示没有图片,初始化时,任意matrix[x,y]当x==0或y==0时,matrix[x,y]=-1 public int[,] matrix { get; set; } public bool testLine(Position current) { //上次连接成功后,第一次点击, if (currentPicture == null) { currentPicture = current; return false; } //连续点击了同一幅图片 if (currentPicture.x==current.x&¤tPicture.y==current.y) { return false; } //不是相同的图像,不能消除 if (current.picture != currentPicture.picture) { lastPicture = null; currentPicture = current; return false; } lastPicture = currentPicture; currentPicture = current; //记录连接路径 List<Position> paths = new List<Position>(); paths.Add(currentPicture); bool finded=false; //向东搜索 finded = _find(paths, 2, Direct.East, lastPicture); if (finded) { return true; } //向南搜索 finded = _find(paths, 2, Direct.South, lastPicture); if (finded) { return true; } //向西搜索 finded = _find(paths, 2, Direct.West, lastPicture); if (finded) { return true; } //向北搜索 finded = _find(paths, 2, Direct.North, lastPicture); if (finded) { return true; } //搜索失败 return false; } /*** * results-> 搜索路径 * chance-> 还剩有转弯的机会,最多2次 * direct-> 当前前进的方向 * target-> 搜索的目标 * **/ protected bool _find(List<Position> results, int chance, Direct direct, Position target) { //路径中不能为空 if (results.Count <= 0) { return false; } //取路径中最后一个 Position currentPos = results.ElementAt(results.Count - 1); //构造下一个将要检测的路径 Position nextPos = new Position(); switch(direct){ case Direct.East: nextPos.x=currentPos.x+1; nextPos.y=currentPos.y; break; case Direct.West: nextPos.x=currentPos.x-1; nextPos.y=currentPos.y; break; case Direct.South: nextPos.x=currentPos.x; nextPos.y=currentPos.y+1; break; case Direct.North: nextPos.x=currentPos.x; nextPos.y=currentPos.y-1; break; } bool overFlow = false; //检测是否已经走出边界 if (nextPos.x < 0 || nextPos.x >= width || nextPos.y < 0 || nextPos.y>=height) { overFlow = true; } if (overFlow) { return false; } //检测是否是目标 if (target.x == nextPos.x&&target.y==nextPos.y) { results.Add(nextPos); return true; } //不是-1 说明该位置有图片,不能通过 if (matrix[nextPos.x, nextPos.y]!=-1) { return false; } //加入路径 results.Add(nextPos); bool find = false; //按照当前方向继续试探 find = _find(results, chance, direct, target); if (find) { return true; } //转弯机会已经耗尽 if (chance <= 0) { //此路不通,将路径最后一个位置删除 results.RemoveAt(results.Count - 1); return false; } //当前方向向东,还可以转弯向南,向北继续试探 if (Direct.East == direct) { find = _find(results, chance-1, Direct.North, target); if (find) { return true; } find = _find(results, chance-1, Direct.South, target); if (find) { return true; } results.RemoveAt(results.Count - 1); return false; } //当前方向向南,还可以转弯向东,向西继续试探 else if (Direct.South == direct) { find = _find(results, chance-1, Direct.East, target); if (find) { return true; } find = _find(results, chance-1, Direct.West, target); if (find) { return true; } results.RemoveAt(results.Count - 1); return false; } //当前方向向西,还可以转弯向南,向北继续试探 else if (Direct.West == direct) { find = _find(results, chance-1, Direct.North, target); if (find) { return true; } find = _find(results, chance-1, Direct.South, target); if (find) { return true; } results.RemoveAt(results.Count - 1); return false; } //当前方向向北,还可以转弯向东,向西继续试探 else { find = _find(results, chance-1, Direct.East, target); if (find) { return true; } find = _find(results, chance-1, Direct.West, target); if (find) { return true; } results.RemoveAt(results.Count - 1); return false; } } //联通成功后,将两个位置设置成空 public void onSuccess() { this.matrix[this.lastPicture.x, this.lastPicture.y] = -1; this.matrix[this.currentPicture.x, this.currentPicture.y] = -1; this.currentPicture = null; this.lastPicture = null; } } public class Position { public int x { get; set; } public int y { get; set; } //picture 表示不同图像,如果是-1,表示为空 public int picture { get; set; } } public enum Direct : int { East = 0, South = 1, West = 2, North = 3 }
请高手多多指正……