Skip to main content

吐血游戏推荐: Braid 时空幻境


朋友推荐的游戏,解谜加动作类。

动作部分有点像超级马里奥,而解谜部分 -- 难死!后期基本就靠攻略了。游戏中充斥着时光倒流等概念,对逻辑,推理有很大考验。

作为一般玩家,到此可能就结束了,而我在游戏中也没感到什么异样,只是觉得音乐有时有些郁闷。

而在Youtube查攻略时发现了有隐藏的星星可以拿,拿到全部星星还有隐藏结局。

星星的拿法可谓变态中的变态,其中某一颗需要2小时才能拿到!!!另外一颗在过了前几关之后就再也不能拿了,必须重新开始,而我发现这个时候已经通了一次全关了。。。

在这种情况下只好让Trainer出动了,于是会漂浮和穿墙了,顺利的又通了一次关,同时再次感叹星星的变态,不仅仅是难度,我第一次通关时竟然一个星星都没看到过 -- 当然看到我也不知道去吃。。。

最后说说剧情,整个游戏是用很多笔记本叙述剧情的,我自己玩的第一遍,感觉无非是救公主,想超级马里奥那种。

但是最后一关,首先一个人抱着公主出现,公主跳上高台高呼救命,而后主人公开始和公主在两条路上奔跑,调整机关,一周到关底,公主回到卧室睡觉,而主人公爬到卧室旁边。

本来,到此就该结束了的。

但是下面时光开始倒流,公主和主人公开始往回跑,直到最后,公主高呼救命,然后一个人(就是一开始那个人,让人以为是魔王之类)将跳下的公主接住,然后救走。

我还觉得有些奇怪,因为这时主人公还在场景里,只能离开。后来看了网上文章才明白,原来主人公是一直在追逐公主,而公主是想躲避甚至杀死主人公。

不得不惊叹这关设计之巧妙!非言语之所及,非亲身体验不可。

而尾声的笔记本透露了更多的剧情,整个游戏其实是倒叙。最后一关是时间上最早发生的,而第一关是结局。在网上看到这句话时,顿时豁然开朗,整个游戏也就变得非常连贯了。

而实际上,根据网上说法,这个游戏是讲了一个关于核弹的故事,主人公是一个研究核弹的科学家,看似还很有道理。但整个游戏丝毫未提核弹二字,实在高明!

整个游戏,不论关卡难度和设计,时间概念,图像, 音乐,以及故事背景和内涵,都堪称极品。不过说实话手感略差一点,但是不影响游戏。

绝对吐血推荐!



参考资料:
http://blog.sina.com.cn/s/blog_565e99380100en6b.html
http://www.mtime.com/my/weiwancheng/blog/1932064/
http://pastangel.blogspot.com/2009/04/braid78.html

Comments

Popular posts from this blog

[转] UTF-8 and Unicode FAQ for Unix/Linux

这几天,这个东西把我搞得很头疼 而且这篇文章好像太大了,blogger自己的发布系统不能发 只好用mail了 //原文 http://www.cl.cam.ac.uk/~mgk25/unicode.html UTF-8 and Unicode FAQ for Unix/Linux by Markus Kuhn This text is a very comprehensive one-stop information resource on how you can use Unicode/UTF-8 on POSIX systems (Linux, Unix). You will find here both introductory information for every user, as well as detailed references for the experienced developer. Unicode has started to replace ASCII, ISO 8859 and EUC at all levels. It enables users to handle not only practically any script and language used on this planet, it also supports a comprehensive set of mathematical and technical symbols to simplify scientific information exchange. With the UTF-8 encoding, Unicode can be used in a convenient and backwards compatible way in environments that were designed entirely around ASCII, like Unix. UTF-8 is the way in which Unicode is used under Unix, Linux, and similar systems. It is now time to make sure that you are well familiar ...

Determine Perspective Lines With Off-page Vanishing Point

In perspective drawing, a vanishing point represents a group of parallel lines, in other words, a direction. For any point on the paper, if we want a line towards the same direction (in the 3d space), we simply draw a line through it and the vanishing point. But sometimes the vanishing point is too far away, such that it is outside the paper/canvas. In this example, we have a point P and two perspective lines L1 and L2. The vanishing point VP is naturally the intersection of L1 and L2. The task is to draw a line through P and VP, without having VP on the paper. I am aware of a few traditional solutions: 1. Use extra pieces of paper such that we can extend L1 and L2 until we see VP. 2. Draw everything in a smaller scale, such that we can see both P and VP on the paper. Draw the line and scale everything back. 3. Draw a perspective grid using the Brewer Method. #1 and #2 might be quite practical. #3 may not guarantee a solution, unless we can measure distances/p...

Moving Items Along Bezier Curves with CSS Animation (Part 2: Time Warp)

This is a follow-up of my earlier article.  I realized that there is another way of achieving the same effect. This article has lots of nice examples and explanations, the basic idea is to make very simple @keyframe rules, usually just a linear movement, then use timing function to distort the time, such that the motion path becomes the desired curve. I'd like to call it the "time warp" hack. Demo See the Pen Interactive cubic Bezier curve + CSS animation by Lu Wang ( @coolwanglu ) on CodePen . How does it work? Recall that a cubic Bezier curve is defined by this formula : \[B(t) = (1-t)^3P_0+3(1-t)^2tP_1+3(1-t)t^2P_2+t^3P_3,\ 0 \le t \le 1.\] In the 2D case, \(B(t)\) has two coordinates, \(x(t)\) and \(y(t)\). Define \(x_i\) to the be x coordinate of \(P_i\), then we have: \[x(t) = (1-t)^3x_0+3(1-t)^2tx_1+3(1-t)t^2x_2+t^3x_3,\ 0 \le t \le 1.\] So, for our animated element, we want to make sure that the x coordiante (i.e. the "left" CSS property) is \(...