第一次写与仙人掌有关的题目,虽然不是动态XXX
这个题目很有趣,问你有多少个同构的仙人掌
首先,仙人掌的问题肯定是要转换成树的问题的,那么可以先找出所以的环,对于每个环新建一个点,向所以这个环上的点连边,把环的边去掉,为了方便,在其他树边上新建一个中点,好,这样就转换成了树,并且等价。
现在考虑要找一个点它是不动的,显然是直径的中心,它不可能和其他的点换,而且新构的树中心是唯一的。把它拎起来,变成有根树。
剩下的考虑一个点,如果它不是环上的,那么如果有K个子树完全相同,那么有K!种方案。怎么判树相同?可以sort一下暴力hash,从小到大标号。
如果是环的中心,那么它不能随便交换,要考虑环上轮换(非中心不用考虑),或者对称,这个是字符串的基础问题了。
然后要考虑如果中心是环的中心,要考虑轮换
注意:
扫描环的时候要考虑顺序
偶环的对称有两种情况,都可以转换成回文串问题
2022年1月25日 21:21
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