你上面那个图不合乎规格阿，缺了5个个直角三角形。按照规则，匹配上一个直角的图形就是矩形的一角，所有能够匹配直角的图形不能超过4个。
  另外你的原题中：
  If pairs of consecutive *s on a common horizontal edge of a piece are stored in columns c1
and c2 , then jc1 􀀀 c2j = 2.
 If pairs of consecutive *s on a common vertical edge of a piece are stored in rows r1 and r2,
then jr1 􀀀 r2j = 1.
 If pairs of consecutive *s on a common diagonal edge of a piece are stored in rows r1 and r2
and in columns c1 and c2 , then jr1 􀀀 r2j = 1 and jc1 􀀀 c2j = 1.
 No piece contains any hole.

 Say that a * stored in row r1 and column c1 is a neighbouring * of a * stored in row r2 and
column c2 if either c1 = c2 and jr1 􀀀 r2j = 1, or r1 = r2 and jc1 􀀀 c2j = 2, or jr1 􀀀 r2j = 1
and jc1 􀀀 c2j = 1. The vertices of a given piece can be identied as those points marked
with a * from which it is possible to reach neighbouring *s in exactly two derent directions
and in no opposite direction.

  这段将边规则的，里面好多乱码，给个清楚点的吧。你那个关于是不是合格三角形的问题，答案可能估计就在里面。

发了题目之后发现没有反应就一直没有来。。。结果发现大家还是注意到了。。。

题目条件说了角不会被切开
还有顶点不会有3邻居（或以上）
还有直角都会被保留，
还有，如果长和宽至少一个是只有两块碎片就可以拼成了，

所以只要找到外部直角（注意：内部也有直角，看输入文件3），然后把直角边相加，上边和上边加，下边加下边，左右也是这样，然后比较长的就是拼成的矩形边了


所以题目就简化了很多

刚拿到手时觉得比较困难，仔细读题目了之后还是发现不难实现的

比较难的面积计算有人用海伦公式，不过我认为我之前说的矢量算法应该是最简单的了

我用数组做的，代码长度200多行