关于《推箱子》游戏用flash实现的关键技术
点击次数：18 次 发布日期：2008-11-21 23:34:15 作者：源代码网
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这是游戏中某一关的示意图：  首先要包含：箱子、推箱子者、箱子可移动的区域、箱子要到达的目的地，这几个基本要素。 我觉得关键要解决的几点就是： 1.在程序中描述可移动区域的边界，并判断箱子和推动者是否已到达了边界。 2.判断推箱子者是否已经碰到了箱子。 3.使推箱子者移动的同时让挨着它的箱子一起移动（完成推动效果）。 4.当箱子被推到了指定地点时和再次被推离指定地点时要做出响应（全部到位是即过关）。 5.记录操作者前三步操作所移动的目标坐标，以供玩家可以及时撤销前一次失误的操作。 ==本人总结了实现的关键技术就是数组地图问题== [布局示意图]  [至于那些墙和箱子等等可以任意画] 首先，在游戏开始时在一关键帧中书写一下代码： // 定义变量 好，至此大功告成，剩下的就是添加四个按钮，用来监测按下“上、下、左、右”时根据不同状态做出判断、移动就OK了。// 已到达目标数、步数、分数、时间、过关需要到位的箱子数、关数 onit = 0; step = 0; time = 60; win = 3; guan = 1; // 主角所在的数组地图位置 mx = 2; my = 1; // 3个箱子分别所在的数组地图位置 // 箱子排列顺序按照：先上后下、先左后右 b1x = 3; b1y = 2; b2x = 2; b2y = 5; b3x = 5; b3y = 6; // 设置数组地图（0 空、1 墙、2 目的地、3 箱子、4 已经就位的箱子） // 数组地图排列规则：a[位于第几列][位于第几行]！（靠！tmd和常理有些区别，节选自《网易动画站》2001年作品，这帮变态！） // 创建二维数组 a = new Array(); i = 0; while (7>=i) { a[i] = new Array("0", "0", "0", "0", "0", "0", "0", "0"); i++; } // 勾勒地图边界 i = 0; while (7>=i) { a[i][0] = "1"; a[i][7] = "1"; a[0][i] = "1"; a[7][i] = "1"; i++; } // 描绘地图形态（0 空、1 墙、2 目的地、3 箱子、4 已经就位的箱子） a[1][1] = "1"; a[1][2] = "1"; a[1][3] = "1"; a[1][4] = "2"; a[1][5] = "2"; a[1][6] = "2"; a[2][1] = "0"; a[2][2] = "0"; a[2][3] = "1"; a[2][4] = "1"; a[2][5] = "3"; a[2][6] = "0"; a[3][1] = "0"; a[3][2] = "3"; a[3][3] = "0"; a[3][4] = "0"; a[3][5] = "0"; a[3][6] = "0"; a[4][1] = "1"; a[4][2] = "0"; a[4][3] = "1"; a[4][4] = "1"; a[4][5] = "0"; a[4][6] = "0"; a[5][1] = "1"; a[5][2] = "0"; a[5][3] = "0"; a[5][4] = "0"; a[5][5] = "1"; a[5][6] = "3"; a[6][1] = "1"; a[6][2] = "1"; a[6][3] = "0"; a[6][4] = "0"; a[6][5] = "0"; a[6][6] = "0"; man_x = getProperty(man, _x); man_y = getProperty(man, _y); box1_x = getProperty(box1, _x); box1_y = getProperty(box1, _y); box2_x = getProperty(box2, _x); box2_y = getProperty(box2, _y); box3_x = getProperty(box3, _x); box3_y = getProperty(box3, _y); 例如按下“右”键后的处理流程应该是： on (keyPress "<Right>") { if ((a[mx + 1][my] == "0") || (a[mx + 1][my] == "2")) { setProperty("man", _x , getProperty(man, _x) + 30); mx = mx + 1; step = step + 1; } else if (a[mx + 1][my] == "3") { if (a[mx + 2][my] == "0") { a[mx + 1][my] = "0"; a[mx + 2][my] = "3"; setProperty("man", _x , getProperty(man, _x) + 30); mx = mx + 1; if ((b1x == mx) && (b1y == my)) { setProperty("box1", _x , getProperty(box1, _x) + 30); b1x = b1x + 1; } if ((b2x == mx) && (b2y == my)) { setProperty("box2", _x , getProperty(box2, _x) + 30); b2x = b2x + 1; } if ((b3x == mx) && (b3y == my)) { setProperty("box3", _x , getProperty(box3, _x) + 30); b3x = b3x + 1; } step = step + 1; } else if (a[mx + 2][my] == "2") { a[mx + 1][my] = "0"; a[mx + 2][my] = "4"; setProperty("man", _x , getProperty(man, _x) + 30); mx = mx + 1; if ((b1x == mx) && (b1y == my)) { setProperty("box1", _x , getProperty(box1, _x) + 30); b1x = b1x + 1; } if ((b2x == mx) && (b2y == my)) { setProperty("box2", _x , getProperty(box2, _x) + 30); b2x = b2x + 1; } if ((b3x == mx) && (b3y == my)) { setProperty("box3", _x , getProperty(box3, _x) + 30); b3x = b3x + 1; } onit = onit + 1; step = step + 1; } } else if (a[mx + 1][my] == "4") { if (a[mx + 2][my] == "0") { a[mx + 1][my] = "2"; a[mx + 2][my] = "3"; setProperty("man", _x , getProperty(man, _x) + 30); mx = mx + 1; if ((b1x == mx) && (b1y == my)) { setProperty("box1", _x , getProperty(box1, _x) + 30); b1x = b1x + 1; } if ((b2x == mx) && (b2y == my)) { setProperty("box2", _x , getProperty(box2, _x) + 30); b2x = b2x + 1; } if ((b3x == mx) && (b3y == my)) { setProperty("box3", _x , getProperty(box3, _x) + 30); b3x = b3x + 1; } onit = onit - 1; step = step + 1; } else if (a[mx + 2][my] == "2") { a[mx + 1][my] = "2"; a[mx + 2][my] = "4"; setProperty("man", _x , getProperty(man, _x) + 30); mx = mx + 1; if ((b1x == mx) && (b1y == my)) { setProperty("box1", _x , getProperty(box1, _x) + 30); b1x = b1x + 1; } if ((b2x == mx) && (b2y == my)) { setProperty("box2", _x , getProperty(box2, _x) + 30); b2x = b2x + 1; } if ((b3x == mx) && (b3y == my)) { setProperty("box3", _x , getProperty(box3, _x) + 30); b3x = b3x + 1; } step = step + 1; } } }  至此，基本功能就可以实现了，至于怎么控制过关，和过了关的动画效果就随便啦。源代码网供稿. |
