, a few years ago I sent a lot of neighborhoods on csdn posts , such as Shanghai Bliss in mind that fruit machine development.
good nonsense stopped, and if you do , and I was the same state , then I am willing to share my experiences to everyone and everyone learning exchanges Here is the source code to achieve
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using System.Collections;//ArrayList 命名空间
using System.Threading;
namespace 连连看
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private Bitmap 图集;//各种图案的图片集合
private int 边长 = 50;//各种图案方块的边长
private int 尺寸 = 10;//布局大小，行数与列数
private bool 已选 = false;//布尔值是否已经选中状态
private int x1, y1, x2, y2;//被选中两个图的坐标
public Point z1, z2;//折点坐标
private int 行 = 10, 列 = 10;//行列间距
private int[] 地图 = new int[12 * 12];//一维数组一百个字节的内存容量
public sbyte 刷新 = new sbyte();
private int 已消除 = 25;//用来给以消除的方块赋值-1来记录以消除的方块为哪些
public int a1, a2, b1, b2;//用来消除电脑查找留下的白框 存放临时两点坐标
public sbyte 是否胜利 = 0;//用来储存成功消除队数 日后用来判断是否获胜
public enum 连线方式 { 直连, 一折, 二折 };//枚举三种连线方式
连线方式 啥方式呀=new 连线方式();//实例化一个lxfs
private void 连连看_Load(object sender, EventArgs e)
{
初始化游戏();
this.CenterToScreen();
}
private void 初始化游戏()//游戏虚拟地图初始化方法
{
刷新 = 3;
this.pictureBox1.Cursor = Cursors.Hand;
this.pictureBox1.Height = 边长 * (行 + 2);
//设置图片框控件初始高度 由于图形矩阵为10*10 加之上下间隙
//间隙算边长所以上下一共算两个间隙单位
this.pictureBox1.Width = 边长 * (列 + 2);
this.Height = this.pictureBox1.Height +
(this.Height - this.ClientRectangle.Height) +
this.menuStrip1.Height;
/*这里this就是"连连看.Form1"也就是设置窗体的高度 窗体的高度
是根据图片框的高度来定的 所以由图片框高度 加上标题栏和菜单栏
的高度 得出的窗体高度*/
this.Width = this.pictureBox1.Width +
(this.Width - this.ClientRectangle.Width);
//窗体宽度基本就是图框宽度了
抽象排列();
生成界面();
}
private void 抽象排列()
{
Random 随机 = new Random();//声明一个随即对象 r
ArrayList 动态数组 = new ArrayList();//声明一个动态数组
for (int i = 0; i < (行 * 列) / 4; i++)
{
for (int j = 0; j < 4; j++)
{
动态数组.Add(i);//动态数组添加4组0到24共计100个ID
}
}
for (int i = 0; i < (行 * 列); i++)
{
int nIndex = 随机.Next() % 动态数组.Count;
地图[i] = (int)动态数组[nIndex];
动态数组.RemoveAt(nIndex);
//至此 数组"地图[i] 内字符以 4组 0~24 不规则随机排列"
//listView1.Items.Add(nIndex.ToString());
}
}
private void 生成界面()//用被截图排矩阵形成初始化图像
{
Graphics g = 界面();
for (int i = 0; i < 100; i++)//最基本的for循环不解释了
{
g.DrawImage(截图(地图[i]), 边长 * (i % 尺寸) +
边长, 边长*(i / 尺寸) + 边长, 边长, 边长);
//画出10乘10的方图矩阵
/*由于数组地图[i]已完成4组 0~24 不规则随机排列,故而以"i"
作为下标来完成截图可以保证所截相同图的ID一致*/
//Graphics g = this.pictureBox1.CreateGraphics();
Pen p = new Pen(Color.Black, 3);
Rectangle a = new Rectangle((i % 尺寸) * 边长 + 边长, (i / 尺寸) * 边长 + 边长, 边长 - 3, 边长 - 3);
g.DrawRectangle(p, a);
}
}
private Graphics 界面()
//此方法以pictureBox控件的背景作为GDI绘图的画板。方法返回一个Graphics类型
{
if(pictureBox1.Image == null)
{
Bitmap bmp = new Bitmap(pictureBox1.Width,pictureBox1.Height);
pictureBox1.Image = bmp;
}
Graphics g = Graphics.FromImage(pictureBox1.Image);
return g;
}
private Bitmap 截图(int 标号)//截图方法
{
//string 图片路径 = 连连看.Properties.Resources.tp;
//图集 = (Bitmap)Image.FromFile(Application.StartupPath + @"\Image\tp.jpg");
//图片实例化并且指定图片集合的路径
图集 = 连连看.Properties.Resources.tp;
//这里图集是 Bitmap类型 而资源库里的图片也是此类型可以直接引用
Bitmap 被截图 = new Bitmap(边长,边长);
//从Bitmap类库中 申明一个 被截图对象 实例化 边长为50
Graphics 图片 = Graphics.FromImage(被截图);
//GDI绘图 用的是所截图 也就是先前申明的 被截图
Rectangle a = new Rectangle(0,0,边长,边长);
//矩形对象 x,y位置 宽高 四个参数
Rectangle b = new Rectangle(标号 * 64, 0, 64, 64);
/*这里用"标号"来控制所截图的不同，比如当标号=2的时候
这里的截图开始的点就变成78了 就变成第二张图了
前两个也是位置坐标参数 后两个是像素 大小 分别是宽与高
*/
图片.DrawImage(图集,a,b,GraphicsUnit.Pixel);
/*截图方法 图集自然就是要截图的图源
a 指定所绘制图像的位置和大小，将图像进行缩放以适合该矩形。
b 图集对象中要绘制的部分。最后 指定参数所用的度量单位。
总的来说 次方法就把图片一块一块截下来 指定边长为长度的正方形
*/
return 被截图;
//方法返回值为bitmap类型 所以返回"被截图"便于之后作为bitmap类
//用于其他方法的参数中
}
private bool Y直连(int y1, int y2, int x)
{
if (y1 > y2)//判断y坐标大小
{
int a = y1;
y1 = y2;
y2 = a;
//如果y1小于y2的则两点换位，用一个a来做中介传递参数，应该都看得懂吧
}
//当第一点的y坐标大于第二点y坐标时，第一点在第二点之下
//循环就要用"i = y1 - 1; i > y2; i--"来完成，为了程序的代码可以重复调用
//所以给两点坐标换位来达到第一点y轴坐标永远小于第二点y轴坐标
//这样一来 "i = y1 + 1; i < y2; i++"只要写一次便可以多次调用，
//写代码就简单了 一劳永逸啊。
for (int i = y1 + 1; i < y2; i++)//这里的代码和x直连方法一样，只是方向不同而已
{
if(i==y2)
{
break;
}
if(地图[i*10+x]!=已消除)
{
return false;
}
}
return true;
}
private bool X直连(int x1, int x2,int y)
{
if (x1 > x2)
{
int n = x1;
x1 = x2;
x2 = n;
}
//x轴方法 三个参数 分别是 第一、二两点的x轴坐标和一个y轴坐标，
//因为y轴相同所以y无所谓那个点
for (int i = x1 + 1; i < x2; i++)
{
//这里循环从第一点的后边一个点开始知道第二点的左边一点
if (i == x2)//如果两点紧挨着则直接跳出循环
{
break;
}
if (地图[y * 10 + i] != 已消除)
//这里用"-1"给数组赋值来记录方块以消除
//判断数组是否有-1的值也就是说方块是否消
//除如果没有消除也就无法连通
{
return false;//则返回，返回"假"值
}
}
return true;//不满足以上条件的话则返回"真"值
}
------ Solution ------------------------------------- -------
good good good good
------ Solution -------------------------- ------------------
good support originality !
------ Solution - -------------------------------------------
Yes, yes, Thank you to share
------ Solution -------------------------------------- ------
support the original, the proposed increase precision .
------ Solution ---------------------------------------- ----
this language so powerful, supporting Chinese variables and functions , exposure. . .
------ Solution ---------------------------------------- ----
teachable teachable
------ Solution ----------------------------- ---------------
support ...
------ Solution ------------------ --------------------------
support principles , MARK
------ Solution ------------------------------------ --------
amount turned out to be Chinese programming embarrassing rz you is not easy to learn language drops before ...
all very well saying that the program code to learn the
------ Solution ------------------------ --------------------
best to package the project directly to csdn the resources for them to learn thank AC
------ Solution ------------------------------------------ -
really take no spare time ah. Can not see such things . Even though I was a test machine , but also from the actual project to find out small problem.
------ Solution ---------------------------------------- ----
looks pretty good
------ Solution ----------------------------- ---------------
very powerful ah , admire
------ Solution ----------------- ---------------------------
cattle ~ ~ ~ ~
------ Solution ----- ---------------------------------------
this good ha ......
------ Solution ------------------------------------------- -
cattle fork, Moo .
------ Solution ---------------------------------------- ----
Bitmap screenshot of the first and the second sentence is a comment ? Careless wrong , right ?
------ Solution ---------------------------------------- ----
collection , also studied . Very good
------ Solution ------------------------------------ --------
decisive collection . . Not bad ;
------ Solution ------------------------------------- -------
are 201X , do not support the Chinese variables and functions behind the bar too
just use Chinese variables and functions , but not the keyword Chinese , generally not Chinese programming
------ For reference only --------------- ------------------------
private bool 一折连(int x1, int y1, int x2, int y2)//联通形式为一个折弯的方法
{
if (y1 < y2)
{
if (地图[y1 * 列 + x2] == 已消除)//右上 ↗
{
if (X直连(x1, x2, y1))
{
if (Y直连(y1, y2, x2))
{
z1.X = x2;
z1.Y = y1;
return true;
}
}
}
if (地图[y2 * 列 + x1] == 已消除)//左下↙
{
if (X直连(x1, x2, y2))
{
if (Y直连(y1, y2, x1))
{
z1.X = x1;
z1.Y = y2;
return true;
}
}
}
}
else
{
if (地图[y2 * 列 + x1] == 已消除)//左上↖
{
if (X直连(x1, x2, y2))
{
if (Y直连(y1, y2, x1))
{
z1.X = x1;
z1.Y = y2;
return true;
}
}
}
if (地图[y1 * 列 + x2] == 已消除)//右下↘
{
if (X直连(x1, x2, y1))
{
if (Y直连(y1, y2, x2))
{
z1.X = x2;
z1.Y = y1;
return true;
}
}
}
}
return false;
}
private bool 右边(int x1, int y1, int x2, int y2)//右边 →
{
for (int x = x1 + 1; x <= 列; x++)
{
if (X直连(x1, x, y1) == true &&
地图[y1 * 尺寸 + x] == 已消除 &&
一折连(x, y1, x2, y2) == true)
{
z1.X = x;
z1.Y = y1;
z2.X = x;
z2.Y = y2;
return true;
}
}
if (X直连(x2, 列, y2))
{
if (X直连(x1, 列, y1))
{
z1.X = 列;
z1.Y = y1;
z2.X = 列;
z2.Y = y2;
return true;
}
}
return false;
}
private bool 左边(int x1, int y1, int x2, int y2)
{
for (int x = 0; x < x1; x++)
{
if (X直连(x1, x, y1) == true &&
地图[y1 * 尺寸 + x] == 已消除 &&
一折连(x, y1, x2, y2) == true)
{
z1.X = x;
z1.Y = y1;
z2.X = x;
z2.Y = y2;
return true;
}
}
if (X直连(x1, -1, y1))
{
if (X直连(x2, -1, y2))
{
z1.X = -1;
z1.Y = y1;
z2.X = -1;
z2.Y = y2;
return true;
}
}
return false;
}
private bool 上边(int x1, int y1, int x2, int y2)
{
for (int y = 0; y <y1; y++)
{
if (Y直连(y, y1, x1) == true &&
一折连(x1, y, x2, y2) == true &&
地图[y * 尺寸 + x1] == 已消除)
{
z1.X = x1;
z1.Y = y;
z2.X = x2;
z2.Y = y;
return true;
}
}
if(Y直连(y1,-1,x1))
{
if (Y直连(y2, -1, x2))
{
z1.X = x1;
z1.Y = -1;
z2.X = x2;
z2.Y = -1;
return true;
}
}
return false;
}
private bool 下边(int x1, int y1, int x2, int y2)
{
for (int y = y1 + 1; y <= 行; y++)
{
if (Y直连(y, y1, x1) == true &&
一折连(x1, y, x2, y2) == true &&
地图[y * 尺寸 + x1] == 已消除)
{
z1.X = x1;
z1.Y = y;
z2.X = x2;
z2.Y = y;
return true;
}
}
if (Y直连(y1, 行, x1))
{
if (Y直连(y2, 行, x2))
{
z1.X = x1;
z1.Y = 行;
z2.X = x2;
z2.Y = 行;
return true;
}
}
return false;
}
private bool 二折连(int x1, int y1, int x2, int y2)
{
try
{
if (右边(x1, y1, x2, y2) == true)
{
return true;
}
if (上边(x1, y1, x2, y2) == true)
{
return true;
}
if (左边(x1, y1, x2, y2) == true)
{
return true;
}
if (下边(x1, y1, x2, y2) == true)
{
return true;
}
}
catch (Exception ex)
{
MessageBox.Show(ex.Message, "趙趑龍提示：",
MessageBoxButtons.OK, MessageBoxIcon.Information);
//return false;
}
return false;
}
private bool 连通判断()
{
if (x1 == x2)//这里很好理解，当两点x轴相同，表明y轴垂直连接
{
if (Y直连(y1, y2, x1)==true)//两点垂直的话
{
啥方式呀 = 连线方式.直连;//枚举方式为直连
return true;//方法返回一个"真"值
}
}
if (y1 == y2)//同理，x轴水平相连
{
if (X直连(x1, x2, y1)==true)
{
啥方式呀 = 连线方式.直连;
return true;
}
}
if (一折连(x1, y1, x2, y2) == true)//满足一折连条件
{
啥方式呀 = 连线方式.一折;//枚举方式为"一折"
return true;//方法返回"真"
}
if (二折连(x1, y1, x2, y2) == true)//同理满足二折条件
{
啥方式呀 = 连线方式.二折;
return true;
}
else
{
return false;//以上连接条件都不满足说明不可以连接则返回"假"值
}
}
public bool 电脑查找()
{
for (int i = 0; i < 100;i++ )//遍历100个小方块图案
{
if(地图[i]==已消除)//判断有被消除的地方
{
continue;//继续遍历直到找到没有消除的方块为止
}
for (int j = i+1; j < 100;j++ )//如果找到再从之后的一个开始遍历第二个
{
if (地图[i].Equals(地图[j]))//当找到与之前有相同ID 的方块的话
{
x1 = i % 10;
y1 = i / 10;
x2 = j % 10;
y2 = j / 10;
/*
我们来看上面的算法
i是地图的索引值假如i是第36个方块，i余上10得到6就是个位数
i再除以10得到3就是十位数上的数字，这样不就得到x1，y1的坐标了吗
把坐标值赋值给x1、x2、y1、y2给下一步进行判断
*/
if (连通判断() == true)//判断两坐标点是否联通
{
return true;//如果联通返回true
}
}
}
}
return false;//都不满足则返回false
}
------ For reference only -------- -------------------------------
private void 画框(int x, int y, Color 颜色, string 线种)
{
Graphics g = this.pictureBox1.CreateGraphics();//声明图片控件为画板
Pen p = new Pen(颜色, 2);//声明实例化画笔
//p.DashPattern = new float[] { 2, 1 };
if (线种 == "虚线")//如果方法参数写虚线则画笔的参数定义成虚线
{
p.DashStyle = System.Drawing.Drawing2D.DashStyle.Dot;//定义虚线的样式为点
}
Rectangle a = new Rectangle(x * 边长 - 1 + 边长, y * 边长 - 1 + 边长, 边长 - 2, 边长 - 2);
//画长方形 此方法介绍过多次 这里不再赘述
g.DrawRectangle(p, a);//不为虚线的话直接画实线的
}
private void 画线(int x1,int y1, int x2,int y2,Color 颜色,连线方式 什么方式呢)
{
/*
* 首先画线方法的参数 四个坐标分别是第一点和第二点的x，y坐标
* 接着是颜色 为什么要设个颜色参数呢，和画框一样 画线方法还承担着
* 消除画线的功能 通过再次画和背景颜色一样的线来达到消除白线的目的
* 最后一个参数就是枚举 是什么连线方式 就如何画线
*/
Graphics g = this.pictureBox1.CreateGraphics();//实例化画板没什么说头
Pen p = new Pen(颜色, 2);//实例化一个 "笔"颜色先由变量代替 粗细为 2
int x_1 = (x1 + 1) * 边长 + 边长 / 2;
//画线是从中间开始二坐标是从左上角开始，所以还是要进行一下加工
//x_1、x_2、y_1、y_1 分别赋值两点的x y 坐标 而z1x、z2x、z1x、z2y
//分别赋值两个折点坐标
int x_2 = (x2 + 1) * 边长 + 边长 / 2;
int y_1 = (y1 + 1) * 边长 + 边长 / 2;
int y_2 = (y2 + 1) * 边长 + 边长 / 2;
int z1x = (z1.X + 1) * 边长 + 边长 / 2;
int z2x = (z2.X + 1) * 边长 + 边长 / 2;
int z1y = (z1.Y + 1) * 边长 + 边长 / 2;
int z2y = (z2.Y + 1) * 边长 + 边长 / 2;
switch (什么方式呢)//下面用switch循环对号入座 什么方式 什么样的画线
{
case 连线方式.直连://如果直连的话
g.DrawLine(p, x_1, y_1, x_2,y_2);//画一条线 参数分别是画笔和四个坐标
break;//满足条件循环打破
case 连线方式.一折:
g.DrawLine(p, x_1, y_1, z1x, z1y);//一折连的话画两条线 第一点到折点
g.DrawLine(p, z1x, z1y, x_2, y_2);//再从折点到第二点就这么简单
break;
case 连线方式.二折:
g.DrawLine(p, x_1, y_1, z1x, z1y);//二折的话三条线 z1的x，y到z2的
g.DrawLine(p, z1x, z1y, z2x, z2y);//z2的 x，y 直到x2的x，y
g.DrawLine(p, z2x, z2y, x_2, y_2);
break;
default:
return;//然则返回
}
}
private void 消除(int x, int y)
{
Graphics g = this.pictureBox1.CreateGraphics();
SolidBrush 刷子 = new SolidBrush(Color.Black);//刷子的声明于实例化 参数为颜色
Rectangle 矩形 = new Rectangle(x*边长+边长,y*边长+边长,边长,边长);
//同样需要一个矩形来规定刷子刷什么样子的图形
g.FillRectangle(刷子,矩形);//然后刷子涂满矩形 把小方块全部涂黑 达到消除方块的目的
}
private bool 相同(int x1, int y1, int x2, int y2)
{
if (地图[y1 * 行 + x1].Equals(地图[y2 * 行 + x2]))
{
return true;
}
else
{
return false;
}
}
private void 判断胜利()
{
if(是否胜利.Equals(50))//当是否胜利==50的话 表示100个方块全部消除
{
Form2 f2 = new Form2();//窗体实例化
//这里用 消息框 "MessageBox.show()"来提示也可以。我新建了一个form窗体
//为的是可以自定义提示框 做得新颖一点而已
f2.ShowDialog();
//显示窗体 这里 f2.Show()就可以显示窗体而 我用了f2.ShowDialog()方法
//为了让用户先处理当前窗体。
this.Hide();//隐藏原窗体 也可以不用隐藏。
}
}
private void pictureBox1_MouseDown(object sender, MouseEventArgs e)
{
if(e.Button==MouseButtons.Right)
{
}
int x, y;//用来获取鼠标点击后的坐标位置
if (e.Button == MouseButtons.Left)//当鼠标点击的是左键
{
画框(a1, b1, Color.Black, "实线");//调用画框方法 画实线 黑色达到去除白框
画框(a2, b2, Color.Black, "实线");
x = (e.X - 边长) / 边长;//给x赋值
y = (e.Y - 边长) / 边长;//y同样赋值为鼠标点击后的y轴坐标
/*格式化鼠标点击的坐标。也就是说，当鼠标点击一个方位得到
一个X点坐标点，比如 159.86 这时，159.86减去左边留白50得到
109.86再除以边长50得到的是一个整形值"2" 本来109.86除以50
应该是2.1978 但是隐形转换过程中 整形值不保留小数点所以得到
的值为 "2"这样不管鼠标点击的方位是多少，最终得到的是格式化
的整形数值 0到9
*/
if (e.X <= 50 || e.Y <= 50 || e.X >= 550 || e.Y >=550)
{
return;
}
//这里添加一个判断，当鼠标点击范围超出了游戏界面或者非指定的范围
//程序返回到初始点，小于50自然是左、上两边留白地方，大于550同样是
//x、y轴右边与下边界的界外
if (地图[y * 行 + x] == 已消除)
{
return;
}
//这里也同样是一个程序异常判断，假如当鼠标点击的y轴为5 x轴为6
//地图[5乘以10+6]= 地图[56] 则当地图[56]值为-1 表示 点击的地方为
//空的时候 程序返回。这样不管鼠标点哪里只要碰到空方块就等于没点
if (已选 == false)
{
x1 = x;
y1 = y;
//判断是否是第一次选择，因为游戏需要选择两个相同图案的方块才
//能让程序判断是不是可以消除的图案，所以需要选择两次，当处于
//"已选值为false"时候把鼠标x轴值赋值给之前声明的整形变量"x1"
//程序开头已声明此变量《private int x1, y1, x2, y2;//被选中两个图的坐标》
画框(x1, y1, Color.White,"实线");
//给选定的方块画个白色的外框提示玩家 此处调用之前已封装好的"画框"方法
已选 = !已选;//布尔值互斥 当值为true时，改值为false 然后反之。
}
if (已选 == true)
{
x2 = x;//然则把第二次选择的点（与第一次选择的坐标不重复）
y2 = y;//赋值，把鼠标点击的位置格式化坐标赋值给x2，y2
//当"已选为true值"时则表示接下来鼠标点击的坐标为第二次选择方块的坐标
if (x1 == x2 && y1 == y2)//判断当两次点击的位置相同时，第二次选择没有意义
{
return;//则返回 重新选择。
}
画框(x2, y2, Color.White, "实线");//画框方法
if (x1 > x2)//如果先点击的坐标点在右边 换位，换位的重要性之前有提到
{
int n = x1;
x1 = x2;
x2 = n;
n = y1;
y1 = y2;
y2 = n;
}
if (相同(x1, y1, x2, y2)==true && 连通判断()==true)
{
//如果两点ID 相同 并且 可以联通
画线(x1, y1, x2, y2, Color.White, 啥方式呀);//画连线 白色
Thread.Sleep(500);//程序暂停0.5秒
//Thread类在 using System.Threading;命名空间下 using是引用
//如果命名空间没有引用的话 System.Threading.Thread.Sleep();亦可
消除(x1, y1);//方块联通后要消除 调用此方法
消除(x2, y2);//第二个方块也消除
画线(x1, y1, x2, y2, Color.Black, 啥方式呀);//消除连线 就是再画条黑色的
画框(x1, y1, Color.Black, "实线");//画黑框来达到消除白框的目的
画框(x2, y2, Color.Black, "实线");//第二个方块的黑框
地图[y1 * 尺寸 + x1] = 已消除;//此坐标点为 以消除状态 赋值25
地图[y2 * 尺寸 + x2] = 已消除;
生成界面();//重新生成界面 来达到刷新目的
已选 = !已选;//布尔值互斥
是否胜利 += 1;//控制消除方块对数的值 增加一点 表示又多了一对
判断胜利();//每次连线成功后判断是不是获得胜利
label1.Text = "已成功消除 " + 是否胜利 + " 对";
}
else//然则
{
已选 = false;//已选 布尔值回到初始值
画框(x1, y1, Color.Black, "实线");//消除画的选框线
画框(x2, y2, Color.Black, "实线");
return;//最后返回重新来过
}
}
}
}
------ For reference only - --------------------------------------
private void 退出游戏ToolStripMenuItem_Click(object sender, EventArgs e)
{
Application.Exit();//程序完全关闭的代码
}
private void 帮助ToolStripMenuItem_Click(object sender, EventArgs e)
{
//这里可以自由发挥
}
private void 重新游戏ToolStripMenuItem_Click(object sender, EventArgs e)
{
pictureBox1.Invalidate();//刷新图框控件里的图片数据
抽象排列();//打乱小方图ID
生成界面();//按照打乱的ID 截图并且矩阵排列
}
private void 调整布局ToolStripMenuItem_Click(object sender, EventArgs e)
{
try
{
刷新--;
if (刷新.Equals(1))
{
//MessageBox.Show("每次游戏只能刷新布局两次","趙趑龍提示：",
//MessageBoxButtons.OK, MessageBoxIcon.Information);
//return;
调整布局ToolStripMenuItem.Enabled = false;
}
else
{
pictureBox1.Invalidate();
Random 随机 = new Random();//声明一个随即对象 r
ArrayList 动态数组 = new ArrayList();//声明一个动态数组
for (sbyte i = 0; i < (行 * 列); i++)
{
动态数组.Add(地图[i]);
}
for (sbyte i = 0; i < (行 * 列); i++)
{
int 变换 = 随机.Next() % 动态数组.Count;
地图[i] = (int)动态数组[变换];
动态数组.RemoveAt(变换);
}
生成界面();
}
}
catch (Exception ex)
{
MessageBox.Show(ex.Message, "趙趑龍提示：",
MessageBoxButtons.OK, MessageBoxIcon.Information);
}
}
private void 电脑查找ToolStripMenuItem_Click(object sender, EventArgs e)
{
画框(x1, y1, Color.Black, "实线");
if (电脑查找() == true)
{
画框(x1, y1, Color.White, "虚线");
画框(x2, y2, Color.White, "虚线");
a1 = x1;
a2 = x2;
b1 = y1;
b2 = y2;
}
}
}
}
--- --- For reference only ---------------------------------------
run after pictures
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this Chinese variables and methods look good awkward. . .
But support LZ, powerful
------ For reference only ----------------------------- ----------
good stuff ! Learn ! ! !
------ For reference only -------------------------------------- -
worth studying
------ For reference only ------------------------------- --------
Oh eh I look ah ! Thank you exchange .
------ For reference only -------------------------------------- -
variable names used herein Chinese point of view but also to be straight , to facilitate communication and learning . Variable modifiers , etc. can not be Chinese .
------ For reference only -------------------------------------- -
me explain the source of original picture by assigning an absolute path to the program to find the " Atlas" in later tests , it was found picture not the same directory as the program is not attending out of the picture . So I put the pictures directly added to the resource library so put the picture into the program . As I am writing a single line of code comments, so the beginning of the comment is to assign the path of the code comments. Friend you understand yet ?
welcome everyone to interactive learning .
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learn good
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landlord hard ah. . .
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Thank you to share . . .
------ For reference only -------------------------------------- -
landlord novice
developers accustomed to using English to do it, otherwise you will never remain at the novice stage
------ For reference only ------------------ ---------------------
------ For reference only ---------------------------------- -----
name also came oh landlord . Use English Kazakh
------ For reference only ---------------------------------- -----
g.DrawImage ( screenshots ( map [i]), side length * (i% size ) +
side length, edge length * (i / size ) + side length, side length, side length ) ;
g.DrawRectangle (p, a);
landlord trouble to help explain these two g and
Graphics g = Interface ( ) ;
is one do ?
I put the code you write up an error : local variable "g" can not be used before it is declared
------ For reference only ---------------------------------- -----
powerful, I have been a bloodthirsty Ya minute fly butterfly ~ ~ ~
------ For reference only ---------------- -----------------------
I rub , Chinese variables , up,
This encoding is time to switch the United Kingdom ,
not be later changed the name of it.
------ For reference only ---------------------------------- -----
nice post !
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Thank you for sharing
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you should put "Graphics g = interface ( ) " on top, because the code that declares a graphics graphics class and instantiate an object before the object interface ( ) method to get the same type of graph , then had an object before you can use various methods of this object .
------ For reference only -------------------------------------- -
added: follow your way to "Graphics g = interface ( ) ; " onto the bottom, then the above g compiler does not know where it came from of it. You said there will be exceptions it.
------ For reference only -------------------------------------- -
English I switched to a shift very skilled, and if not , then I generally do not have tutorials Chinese do not bother to write a comment . English for the compiler is byte memory only, as long as there had been Chinese vs08 statement it will automatically jump out and sometimes do not play characters .
------ For reference only -------------------------------------- -
explain the very detailed, thank you for sharing
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lz cow ! Studied .
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speaking my former Chinese leader on the special love to use variable methods , can not wait to use Chinese keywords . .
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bunkers . . Too strong
------ For reference only ----------------------------------- ----
- ----- For reference only ---------------------------------------
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good, learned
------ For reference only -------------------------- -------------
Thank you to share . Bookmark it !
------ For reference only -------------------------------------- -
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Thank you to share . Bookmark it !
------ For reference only -------------------------------------- -
test , why, I was also writing lianliankan dota .
------ For reference only -------------------------------------- -
different, completely different, I'm not the same search algorithm . Oh .
------ For reference only -------------------------------------- -
really fuckin to the force !
------ For reference only -------------------------------------- -
Chinese variable and function names , or the first time I saw , support
------ For reference only --------------------- ------------------
legendary Chinese programming
------ For reference only ------------ ---------------------------
support , ------------------ ---
------ For reference only ----------------------------------- ----
good, support it, or have time to , ah
------ For reference only -------------------- -------------------
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taught , the first collection , and take your time. However , I feel identifiers with Chinese still somewhat authentic , is not it ! !
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cattle ah, very clear idea
------ For reference only ------------------------ ---------------
+1
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share your algorithm together and discussing anyway.
------ For reference only -------------------------------------- -
Personally, I think it, for the compiler variable name in English or, Chinese 's worth mentioning, but that is nothing different object -oriented , which is still played the role of memory variables . To echo . NET C # language on the platform object-oriented concepts , I put English into Chinese objects object .
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proposal sent to the code . . . . .
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cattle too ! save !
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saying LZ engage wow pictures
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learn, Thank you so much
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very Good good detail collected
------ For reference only ------------------ ---------------------
thank the landlord
------ For reference only ---------------------------------------
Wow
good strong
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too much of it
------ For reference only ------------------------ ---------------
studied Thank you to share it
------ For reference only ------------- --------------------------
learning , a collection of
------ For reference only ----- ----------------------------------
landlord worth learning , I am sad reminder of work for a year , and feel still did not grow much , technology is still far worse , would like to pay tribute to the landlord !
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