[Flash]flash编程入门，用actionscript3.0实现经典游戏【连连看】

连连看作为一个经典的游戏受到很多上班族的喜爱

我老婆就很喜欢玩这个游戏，好，给她做一个

做起来也不难，只要知道实现的原理，不管用什么语言都很容易实现

先看效果

怎么去实现呢？一下是我的实现原理

1. 加载图片将图片

2. 创造一个二维数组，保存成对的随机图片，并显示在屏幕上

3. 连接算法。

算法是关键，我的思路很简单，首先判断是不是在一条直线上

然后再判断经过两个折线是不是能连接，最后判断经过三条折线是不是能够连接

以下是核心代码

private function SeekRoad():Array
        {
            var road:Array = new Array();
            
            //在一条直线上            
            if(IsThrough(this.firstClick_.ix,this.firstClick_.iy,this.secondClick_.ix,this.secondClick_.iy))
            {
                road.push(GetPathPoint(this.firstClick_.ix,this.firstClick_.iy));
                road.push(GetPathPoint(this.secondClick_.ix,this.secondClick_.iy));
                return road;
            }
            
            for(var i:uint=0;i<GV.WidthNum+2;i++)
            {
                if(this.map_[i][this.firstClick_.iy].Type == 0 && IsXThrough(this.firstClick_.ix,this.firstClick_.iy,i,this.firstClick_.iy))
                {
                    road = new Array();
                    road.push(GetPathPoint(this.firstClick_.ix,this.firstClick_.iy));
                    road.push(GetPathPoint(i,this.firstClick_.iy));
                    //两折
                    if(IsYThrough(i,this.firstClick_.iy,this.secondClick_.ix,this.secondClick_.iy))
                    {
                        road.push(GetPathPoint(this.secondClick_.ix,this.secondClick_.iy));
                        return road;
                    }
                    //三折
                    if(this.map_[i][this.secondClick_.iy].Type == 0 && IsYThrough(i,this.firstClick_.iy,i,this.secondClick_.iy))
                    {
                        road.push(GetPathPoint(i,this.secondClick_.iy));
                        if(IsXThrough(i,this.secondClick_.iy,this.secondClick_.ix,this.secondClick_.iy))
                        {
                            road.push(GetPathPoint(this.secondClick_.ix,this.secondClick_.iy));
                            return road;
                        }
                    }
                }
            }
            
            for(var j:uint=0;j<GV.HeightNum+2;j++)
            {
                if(this.map_[this.firstClick_.ix][j].Type == 0 && IsYThrough(this.firstClick_.ix,this.firstClick_.iy,this.firstClick_.ix,j))
                {
                    road = new Array();
                    road.push(GetPathPoint(this.firstClick_.ix,this.firstClick_.iy));
                    road.push(GetPathPoint(this.firstClick_.ix,j));
                    //两折
                    if(IsXThrough(this.firstClick_.ix,j,this.secondClick_.ix,this.secondClick_.iy))
                    {
                        road.push(GetPathPoint(this.secondClick_.ix,this.secondClick_.iy));
                        return road;
                    }
                    //三折
                    if(this.map_[this.secondClick_.ix][j].Type == 0 && IsXThrough(this.firstClick_.ix,j,this.secondClick_.ix,j))
                    {
                        road.push(GetPathPoint(this.secondClick_.ix,j));
                        if(IsYThrough(this.secondClick_.ix,j,this.secondClick_.ix,this.secondClick_.iy))
                        {
                            road.push(GetPathPoint(this.secondClick_.ix,this.secondClick_.iy));
                            return road;
                        }
                    }
                }
            }
            return new Array();
        }

代码还未优化，仅供参考