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

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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

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