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想法太复杂，用筛法就可以AC了，时间O（n^2）。In Reply To:这个题好象没有人问~不好意思,能不能给个提示! Posted by:helloeveryone at 2004-09-12 22:35:44 每个点(x,y)给个标志1,如果被(x1,y1)覆盖(x1>=x&&y1>=y),则变为0;用每个标志为1的点去覆盖其它点.最后数一下标志为1的点就可以了。如果用链表,时间快了，但代码长了。而写代码是很辛苦的-^- Followed by: Post your reply here: |
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