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由此想到了一个剪枝: 如果一个 match 对 square 的贡献集合, 包含于另一个 match 的贡献集合, 就不用考虑了

Posted by sunmoonstar at 2006-05-16 13:35:05 on Problem 1084
In Reply To:提供一个不用搜索的方法，正确与否尚待诸位检验 Posted by:seol at 2006-05-16 12:54:52 
> 将边与正方形都视为结点，看成二分图(还是叫二部图？偶不知道) 
> 然后在边所构成的结点集合中求最小割，转化为简单的图论问题.
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