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coder114514 committed Dec 2, 2023
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2 changes: 1 addition & 1 deletion _drafts/obsidian-md/blog/2022-11-22-Floyd和倍增Floyd.md
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初始状态
$$
\begin{cases}
f_{0,i,i}=0\
f_{0,i,i}=0\\
f_{0,i,j}=w&如果i到j有一条权值为w的边(重边就选最小的)\\
f_{0,i,j}=+\infty&如果i到j没有边
\end{cases}
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20 changes: 20 additions & 0 deletions _drafts/obsidian-md/blog/2023-12-02-NOIP2023退役记.md
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前几次 [[2022-12-06-NOIP2022游寄]] [[2023-09-20-CSP-S-2023-复赛]]

## 开考
早上起来睡得不错,不怎么困
8:00 多到考场,在七宝中学,他们在修路
进考场以后就是调机器等密码,拿到密码后开始干题目

T1 一眼大水题,半小时之后完成,过了大样例

T2 先写了一个暴力,然后开始考虑正解。发现经过所有的操作之后每个变量要么等于一个常量,要么和另外一个变量相等,要么和另外一个变量的逆相等,有点像 2-SAT。~~忘记 2-SAT 是用并查集做的了~~,于是给每个变量一个点,每个变量逆一个点,每个常量一个点,如果有 $A=B$ 就建一条 $A$ 到 $B$ 的单向边,然后 $\neg x$ 到 $x$ 也有边。发现这是个基环森林,外加每个常量是没有出边的。于是如果 $x$ 能走到 $\neg x$,那么这个分量就必须全是 $U$ 了,再加上直接是常量 $U$ 的就是答案。开心地写好一个程序,第2个样例就假了,于是就开始疯狂修改(碰到地情况包括但不限于输出错误答案,程序卡住等。中途手画了不少图,不过还是有缓慢的进展的。卡在某个样例上(具体哪个忘了,经过大量研究后发现 $x$ 能走到 $\neg x$ 这个条件不对,因为可能中间反转了奇数次,所以一致性还是能满足的。目前为止知道了如果 $x$ 能走到 $\neg x$,则 $x$ 必定在某个环上,以及找出分量遍历环的时候也需要记录走形如 $\neg x$ 到 $x$ 的边的数量。经过2小时的苦干后,过了所有样例。~~(后来知道了并查集做多简单~~

T3 一眼不会做,T4 看最后时间不多了,且暴力感觉不好写,就直接输出了一个样例。

## 后记
看其他人游记,开始担心自己做的 T1 T2 是不是有问题,感觉这次又要寄了

后来出成绩了,100+100+0+0=200,还算不错,算是一个好结局吧。

## 总结
T2 调代码时间用得太多了,T4 其实也不是很难,如果多花点时间,应该还是能干不少分的。
10 changes: 5 additions & 5 deletions index.html
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layout: archive
---

<h2 class="archive__subtitle">链接</h3>
<h3><a href="/obsidian/blog/0000-00-00-目录.html">博客</a></h3>
<h3><a href="/old_blogs/">旧博客</a></h3>
<h3><a href="https://github.com/coder114514/coder114514.github.io">GitHub</a></h3>
<h3><a href="http://coder114514.ysepan.com/">永硕E盘</a></h3>
<h3 class="archive__subtitle">链接</h3>
<h2><a href="/obsidian/blog/0000-00-00-目录.html">博客</a></h3>
<h2><a href="/old_blogs/">旧博客</a></h3>
<h2><a href="https://github.com/coder114514/coder114514.github.io">GitHub</a></h3>
<h2><a href="http://coder114514.ysepan.com/">永硕E盘</a></h3>
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