minesweeper

我在GitHub上找到了这些游戏项目,刺激!

情到浓时终转凉″ 提交于 2021-02-20 11:36:49
作者:JackTian 来源:公众号「杰哥的IT之旅」 ID:Jake_Internet 大家好,我是JackTian。 当你学习、工作累了,趁休息时间娱乐放松一下,通过用游戏的方式来学习技术,那么 GitHub 上这些好玩的开源项目小游戏是最适合不过了。 首先,我们来看一下这张图,可快速了解这篇文章中所涉及到的 23 个关于游戏方面的开源项目。 Games on GitHub 这个开源项目作者收集了托管在 GitHub 上的开源游戏以及跟游戏相关的项目列表,所含的项目类型众多,游戏种类包括教育类、浏览器类等,还包括一些开源的游戏引擎等等; 该项目的目录: 具体细节就不逐一点开给大家演示了,感兴趣的朋友去看看。 GitHub 地址:https://github.com/leereilly/games vim-game-code-break 这个项目是 Vim 插件版本的打豆豆。 将其放在您的.vimrc中: Plug 'johngrib/vim-game-code-break' 然后在Vim中运行以下命令: :source % :PlugInstall 装上这个插件输入 :VimGameCodeBreak ,即可开始游戏。 功能介绍 h、l 键是控制左右方向; space 键是发球; ` 键是作弊键; ] 键是神模式,[ 键是人工模式; q 键是结束游戏,Q 键是退出并关闭游戏;

Python Logic in Minesweeper

好久不见. 提交于 2021-02-08 10:12:55
问题 I am currently writing a Minesweeper game. It was going fine until I encountered the problem with exposing multiple tiles in one click. Whenever I click something, it never shows mines (like it's supposed to) but only ever works its way down, to the left, and to the right. I made it so it prints its direction, and it thinks down is right, etc. I have never encountered a left. Here is my code: from tkinter import * from random import * root = Tk() root.resizable(0, 0) root.title("Minesweeper")

Java programming - nested for loops for minesweeper game

允我心安 提交于 2021-01-28 05:28:44
问题 Currently making a minesweeper game and stuck on being able to loop through the 8 surrounding cells to find the number of surrounding bombs. I have a helper method that I believe works fine that is used to check if the cell (passed by numberOfAdjacentBombs) contains a bomb. How do I go about creating a nested for loop that will loop through each surrounding cell? e.g (row-1, col-1), (row-1, col), (row-1, col+1). Any help or hints is appreciated, thanks! :-) private int numberOfAdjacentBombs

Leetcode之广度优先搜索(BFS)专题-529. 扫雷游戏(Minesweeper)

两盒软妹~` 提交于 2020-11-06 04:44:44
Leetcode之广度优先搜索(BFS)专题-529. 扫雷游戏(Minesweeper) BFS入门详解: Leetcode之广度优先搜索(BFS)专题-429. N叉树的层序遍历(N-ary Tree Level Order Traversal) 让我们一起来玩扫雷游戏! 给定一个代表游戏板的二维字符矩阵。 'M' 代表一个未挖出的地雷,'E' 代表一个未挖出的空方块,'B' 代表没有相邻(上,下,左,右,和所有4个对角线)地雷的已挖出的空白方块,数字('1' 到 '8')表示有多少地雷与这块已挖出的方块相邻,'X' 则表示一个已挖出的地雷。 现在给出在所有未挖出的方块中('M'或者'E')的下一个点击位置(行和列索引),根据以下规则,返回相应位置被点击后对应的面板: 如果一个地雷('M')被挖出,游戏就结束了- 把它改为 'X'。 如果一个没有相邻地雷的空方块('E')被挖出,修改它为('B'),并且所有和其相邻的方块都应该被递归地揭露。 如果一个至少与一个地雷相邻的空方块('E')被挖出,修改它为数字('1'到'8'),表示相邻地雷的数量。 如果在此次点击中,若无更多方块可被揭露,则返回面板。 示例 1: 输入: [['E', 'E', 'E', 'E', 'E'], ['E', 'E', 'M', 'E', 'E'], ['E', 'E', 'E', 'E', 'E'],

1秒内通关扫雷?他创造属于自己的世界记录!Python实现自动扫雷

大城市里の小女人 提交于 2020-05-05 17:58:59
五一劳动节假期,我们一起来玩扫雷吧。用Python+OpenCV实现了自动扫雷,突破世界记录,我们先来看一下效果吧。 中级 - 0.74秒 3BV/S=60.81 相信许多人很早就知道有扫雷这么一款经典的游(显卡测试)戏(软件),更是有不少人曾听说过中国雷圣,也是中国扫雷第一、世界综合排名第二的郭蔚嘉的顶顶大名。扫雷作为一款在Windows9x时代就已经诞生的经典游戏,从过去到现在依然都有着它独特的魅力:快节奏高精准的鼠标操作要求、快速的反应能力、刷新纪录的快感,这些都是扫雷给雷友们带来的、只属于扫雷的独一无二的兴奋点。 ▍准备 准备动手制作一套扫雷自动化软件之前,你需要准备如下一些工具/软件/环境 - 开发环境 Python3 环境 - 推荐3.6或者以上 [更加推荐Anaconda3,以下很多依赖库无需安装] numpy依赖库 [如有Anaconda则无需安装] PIL依赖库 [如有Anaconda则无需安装] opencv-python win32gui、win32api依赖库 支持Python的IDE [可选,如果你能忍受用文本编辑器写程序也可以] - 扫雷软件 · Minesweeper Arbiter 下载地址(必须使用MS-Arbiter来进行扫雷!) 好啦,那么我们的准备工作已经全部完成了!让我们开始吧~ ▍ 实现思路 在去做一件事情之前最重要的是什么

win10 LTSC系统 安装应用商店和纸牌合集,解决从应用商店安装Solitaire Collection纸牌打开空白的问题

痞子三分冷 提交于 2020-04-28 21:39:48
家里台式机换了win10系统,想给老妈玩那个纸牌游戏(我也超喜欢的!。 发现这个系统没有自带纸牌游戏 Microsoft Solitaire Collection , 过分的是,连 应用商店 都没有。。。呵呵呵 于是,我找到了这个教程:http://www.xitongcheng.com/jiaocheng/win10_article_47654.html 打开管理员Windows PowerShell的窗口, 输入命令 Get-AppxPackage -allusers | Select Name, PackageFullName ; 按照他说的,找到 Microsoft.WindowsStore 菜单项, 把其后面的内容复制下来;然后接着再在下面的命令行输入: Add-appxpackage -register "C:\Program Files\WindowsApps\Microsoft.WindowsStore_11701.1001.99.0_x64__8wekyb3d8bbwe\appxmanifest.xml" -disabledevelopmentmode 可怕的是,我竟没有找到所谓的WindowStore,唔。。。这没法继续啊。。。 于是,我找到了这个教程: Windows10 LTSB/LTSC版安装应用商店与UWP-1米以内-51CTO博客 https:/

Python: floodfill algorithm for minesweeper

梦想的初衷 提交于 2020-01-16 18:38:11
问题 I'm writing a minesweeper game in python, and am currently trying to implement a floodfill algorithm. This is what I've written: def floodfill(CURRENT_BOARD, row, col): count = count_mines(row, col) if count in POSSIBLE_MINE_NUMBERS: CURRENT_BOARD[row][col] = str(count) + ' ' else: if CURRENT_BOARD[row][col] == 'P ': CURRENT_BOARD[row][col] = 'P ' if row > 0: floodfill(CURRENT_BOARD, row - 1, col) if row < len(BOARD[row]) - 1: floodfill(CURRENT_BOARD, row + 1, col) if col > 0: floodfill

What is the best way to build a Minesweeper Board in a JFrame?

拜拜、爱过 提交于 2020-01-14 07:02:47
问题 I am just trying to construct a gameboard with tiles to work with. I got it to work by painting the tile into rows and columns with BufferedImage, but I am going to need to constantly update the GUI as the game is played. So I tried to add the tile image to an ImageIcon, then to a JLabel array so that I could set the position. After that I would add the labels to the gui JPanel, but they had these default gaps between them. Am I approaching this incorrectly? I am very new to Swing and Java

What is the best way to build a Minesweeper Board in a JFrame?

China☆狼群 提交于 2020-01-14 07:02:29
问题 I am just trying to construct a gameboard with tiles to work with. I got it to work by painting the tile into rows and columns with BufferedImage, but I am going to need to constantly update the GUI as the game is played. So I tried to add the tile image to an ImageIcon, then to a JLabel array so that I could set the position. After that I would add the labels to the gui JPanel, but they had these default gaps between them. Am I approaching this incorrectly? I am very new to Swing and Java

Generate a minesweeper board which doesn't need guessing

别说谁变了你拦得住时间么 提交于 2020-01-11 16:37:47
问题 I am designing a Minesweeper-like game (with modified rules), and I want to prevent player from guessing. My goal is: The generated board is with few revealed squares, and player can solve the entire puzzle without any guessing. Wikipedia mentioned: Some implementations of Minesweeper will set up the board by never placing a mine on the first square revealed, or by arranging the board so that the solution does not require guessing. However, I cannot figure out the algorithm. Besides, in