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对,就是O(N). 先dfs造一棵树.然后从叶往父结点添加人数.每个结点记录它多带出的子树的人数.对于每个结点,取abs((total-now)-now)就是差值了...然后记录一个minIn Reply To:Re:复杂度只为O(N)，应该不会超时 Posted by:byron at 2006-11-23 12:18:07 Followed by: Post your reply here: |
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