CF232C Doe Graphs】的更多相关文章

传送门 Solution:  (不理解时对着图研究一下就清楚啦!!!) sm[i]为|D(i)|  (x,y,n)为x,y在D(n)中的最短路 已知sm[i-1]+1为D(i)的割点 于是x-y的最短路就可以分为三种情况: x<sm[n-1]+1&&y>=sm[n-1]+1 x<sm[n-1]+1&&y<sm[n-1]+1 x>=sm[n-1]+1&&y>=sm[n-1]+1 下面我们就来讨论这三种情况 x在图D(n-1)…
CF232C Doe Graphs 题意 题意翻译 \(Doe\)以她自己的名字来命名下面的无向图 \(D(0)\)是只有一个编号为\(1\)的结点的图. \(D(1)\)是只有两个编号分别为\(1\)和\(2\)的点与一条连接这两个点的边的图. \(D(n)\)以如下方法构造 将\(D(n-2)\)中所有点的编号加上\(|D(n-1)|\) 在点\(|D(n-1)|\)与点\(|D(n-1)|+1\)之间连边. 在点\(|D(n-1)|+1\)与点1之间连边 现在\(Doe\)已经构造出了\(…
由于换了台电脑,而我的贪心 & 构造能力依然很拉跨,所以决定再开一个坑( 前传: 贪心/构造/DP 杂题选做 u1s1 我预感还有Ⅲ(欸,这不是我在多项式Ⅱ中说过的原话吗) 24. P5912 [POI2004]JAS 一开始直接莽了个点分治,当我测过了样例美滋滋地一交,发现自己获得了 20 分的好成绩之后,才发现事情有那么亿点点不对劲( 不难发现,题目等价于求高度最小的点分树的高度,直接求有点困难,我们不妨来对其进行一些转化:我们考虑给每个点一个标号,那么问题可以转化为,求使得任意两个标号相同…
A. Perfect Permutation 奇偶对调. B. Non-square Equation \(s(x)\)不超过200,根据求根公式计算\(x\). C. Cycles 每次新增点时都和前面的点连边,即每次增加\(i\)个点,如果\(k<i\)则需要新增加点. D. Table 如果前\(n\)列每列的个数确定,那么整个表格的每列的个数也对应确定了. 如果第\(i(1 \le i \le n)\)列的个数为\(c_i\),那么\(c_i\)会重复\[1+\frac{m-i}{n}\…
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