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来踩个脚印吧!

Comments (8)

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这个貌似只需逐点验证即可吧,把王周围3格以内情况都列举一遍。不论骑士与王的初始位置如何,最短步数的相遇点一定在2格范围内,因为骑士比王速度快。
Oct. 28
tonywrote:
我想问个问题usaco3.3Camelot,你能不能证明出“显然王走的比骑士要慢,那么应该尽量要让王少走,所以王需要先走的情况只可能是骑士无法到达的地方或者骑士到达需要绕一圈的情况。可以构想一下骑士在王附近时达到王的路径,就不难发现,相遇点只应该在王的附近很短的距离以内。估算一下,相遇点就在王的坐标加减2的范围内枚举就可以了,经过证明,这样做是正确的(好象有人已经在OIBH上发表了证明方法)。加上这两个优化,就可以比较快地AC了。 ”这句话?(摘自nocow)
Oct. 21
新年快乐~
Jan. 25
卡修 本wrote:
最近忙什么不?帮我个小忙吧,俺要做个控件。。。不照了,俺msn:xuchao611@hotmail.com见到速联系我,不然偶就华丽的被老师虐了
June 7
No namewrote:
让我在这里泪奔一下,樱哥你不留言我还找不到这里呢。。。电脑重装了N次。。。很多资料都米了
现在开始考一模了。。。以后再慢慢收拾你,表问WHY。。。= =+
Apr. 21
卡修 本wrote:
加我的msn:xuchao611@hotmail.com,顺便告诉我你的msn我好加你,呵呵,快考研了,好好学习,天天向上
Oct. 16
我知道你的一切资料,可以到你的寝室搞暗杀,呵呵
从哪里看出我也是USTC的啊?
Oct. 12
卡修 本wrote:
你好,呵呵,你也是ustc兄弟?
Oct. 12