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300题留念，谈谈做法正宗做法貌似是线段树，感觉那样做每个更新复杂度为nlogn，而且不太好维护因此直接搞了，每种颜色分开搞，开始先统计出每种颜色方案数。由于某个元素颜色改变只会影响以当前行和当前列的元素为中心的cross，因此只需更新这些点变化后cross数目发生的改变即可 代码：http://shaidaima.com/source/view/11148 Followed by: Post your reply here: |
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