ARC119F 题解

blog。被自动机做法恶心到了,现在也来恶心一下大家。

\(\color{red}\textbf{以下内容强烈建议自己推一遍,几乎一半是重复的,推完会很爽,并且理解会很深。}\)

\(\color{red}\textbf{以下内容不建议用} \LaTeX\textbf{书写,因为写起来像在吃大便。}\)


暴力 \(dp_{i,a,b}\) 表示当前在 \(i\),最近的 A/B 在 \(a/b\),方案数。这个是三次方的。

考虑优化状态:比如说 \(\texttt{BAAAAA...AAB}\) 当前在 B,显然傻逼都知道跳过去最优,那么中间无论有几个 A,state 都是相同的,可以尝试记为同一个 state。

于是直接暴力枚举转移,可以实现一个自动机。

提取大便

开始吃大便!下面标绿的是状态,标红的是当前所在处。

  • \(\color{green}\texttt{A}\)\(\texttt{B}\color{red}\texttt{A}\),意为当前在 A,前一个是 B,后面的不重要。
  • \(\color{green}\texttt{B}\)\(\texttt{A}\color{red}\texttt{B}\)
  • \(\color{green}\texttt{AA}\)\(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{A}\)
  • \(\color{green}\texttt{BB}\)\(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{B}\)
  • \(\color{green}\texttt{AAA}\)\(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AA}\)
  • \(\color{green}\texttt{BBB}\)\(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BB}\)
  • \(\color{green}\texttt{AAAA}\)\(\color{red}\texttt{B}\color{black}\texttt{AAA..AA}\)。注意即使 \(cnt_A<4\) 但是从 \(\texttt{B}\) 直接跳最优的话,也归属于这个 state。
  • \(\color{green}\texttt{BBBB}\)\(\color{red}\texttt{A}\color{black}\texttt{BBB..BB}\)。同上。
  • \(\color{green}\texttt{AB}\)\(\texttt{?}\color{red}\texttt{A}\color{black}\texttt{B}\)
  • \(\color{green}\texttt{BA}\)\(\texttt{?}\color{red}\texttt{B}\color{black}\texttt{A}\)
  • \(\color{green}\texttt{@}\)\(\color{red}\texttt{?}\)。注意这里问号的意义是「既可以当作 A 也可以当作 B」。
  • \(\color{green}\texttt{@A}\)\(\color{red}\texttt{?}\color{black}\texttt{A}\)
  • \(\color{green}\texttt{@B}\)\(\color{red}\texttt{?}\color{black}\texttt{B}\)

品尝大便

总计 \(13\) 个状态。转移考虑添加 \(\texttt{A/B}\),然后尝试用最小代价跑到定义过的状态去。

对于状态 \(\color{green}\texttt{A}\)\(\color{green}\texttt{B}\),我们有:

  • \(\texttt{B}\color{red}\texttt{A}\color{black}\to\texttt{B}\color{red}\texttt{A}\color{black}\texttt{A}\):不用走即有状态 \(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{A}\)\(\color{green}\texttt{A}\color{black}\xrightarrow{0}\color{green}\texttt{AA}\color{black}\ (\text{add A})\)
  • \(\texttt{B}\color{red}\texttt{A}\color{black}\to\texttt{B}\color{red}\texttt{A}\color{black}\texttt{B}\):不用走即有状态 \(\color{darkgrey}\texttt{B}\color{red}\texttt{A}\color{black}\texttt{B}\)\(\color{green}\texttt{A}\color{black}\xrightarrow{0}\color{green}\texttt{AB}\color{black}\ (\text{add B})\)
  • \(\texttt{A}\color{red}\texttt{B}\color{black}\to\texttt{A}\color{red}\texttt{B}\color{black}\texttt{A}\):不用走即有状态 \(\color{darkgrey}\texttt{A}\color{red}\texttt{B}\color{black}\texttt{A}\)\(\color{green}\texttt{B}\color{black}\xrightarrow{0}\color{green}\texttt{BA}\color{black}\ (\text{add A})\)
  • \(\texttt{A}\color{red}\texttt{B}\color{black}\to\texttt{A}\color{red}\texttt{B}\color{black}\texttt{B}\):不用走即有状态 \(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{B}\)\(\color{green}\texttt{B}\color{black}\xrightarrow{0}\color{green}\texttt{BB}\color{black}\ (\text{add B})\)

对于状态 \(\color{green}\texttt{AA}\)\(\color{green}\texttt{BB}\),我们有:

  • \(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{A}\to\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AA}\):不用走即有状态 \(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AA}\)\(\color{green}\texttt{AA}\color{black}\xrightarrow{0}\color{green}\texttt{AAA}\color{black}\ (\text{add A})\)
  • \(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{A}\to\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AB}\):往后走一步即有状态 \(\color{grey}\texttt{BA}\color{red}\texttt{A}\color{black}\texttt{B}\)\(\color{green}\texttt{AA}\color{black}\xrightarrow{1}\color{green}\texttt{AB}\color{black}\ (\text{add B})\)
  • \(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{B}\to\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BA}\):往后走一步即有状态 \(\color{grey}\texttt{AB}\color{red}\texttt{B}\color{black}\texttt{A}\)\(\color{green}\texttt{BB}\color{black}\xrightarrow{1}\color{green}\texttt{BA}\color{black}\ (\text{add A})\)
  • \(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{B}\to\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BB}\):不用走即有状态 \(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BB}\)\(\color{green}\texttt{BB}\color{black}\xrightarrow{0}\color{green}\texttt{BBB}\color{black}\ (\text{add B})\)

对于状态 \(\color{green}\texttt{AAA}\)\(\color{green}\texttt{BBB}\),我们有:

  • \(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AA}\to\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AAA}\),往前走一步即有状态 \(\color{red}\texttt{B}\color{black}\texttt{AAAA}\)\(\color{green}\texttt{AAA}\color{black}\xrightarrow{1}\color{green}\texttt{AAAA}\color{black}\ (\text{add A})\)
  • \(\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AA}\to\texttt{B}\color{red}\texttt{A}\color{black}\texttt{AAB}\),发现走一步啥都到不了,而走两步可以到 \(\color{grey}\texttt{BAA}\color{red}\texttt{A}\color{grey}\texttt{B}\)(往后两步)或 \(\color{grey}\texttt{BAAA}\color{red}\texttt{B}\)(往前一步然后跳过去),此时同时可得两个状态 \(\color{green}\texttt{A}\)\(\color{green}\texttt{B}\),故 \(\color{green}\texttt{AAA}\color{black}\xrightarrow{2}\color{green}\texttt{@}\color{black}\ (\text{add B})\)
  • \(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BB}\to\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BBA}\),同上,\(\color{green}\texttt{BBB}\color{black}\xrightarrow{2}\color{green}\texttt{@}\color{black}\ (\text{add A})\)
  • \(\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BB}\to\texttt{A}\color{red}\texttt{B}\color{black}\texttt{BBB}\),往前走一步即有状态 \(\color{red}\texttt{A}\color{black}\texttt{BBBB}\)\(\color{green}\texttt{BBB}\color{black}\xrightarrow{1}\color{green}\texttt{BBBB}\color{black}\ (\text{add B})\)

对于状态 \(\color{green}\texttt{AAAA}\)\(\color{green}\texttt{BBBB}\),我们有:

  • \(\color{red}\texttt{B}\color{black}\texttt{AAA..AA}\to\color{red}\texttt{B}\color{black}\texttt{AAA..AAA}\),不用动即为原状态,\(\color{green}\texttt{AAAA}\color{black}\xrightarrow{0}\color{green}\texttt{AAAA}\color{black}\ (\text{add A})\)
  • \(\color{red}\texttt{B}\color{black}\texttt{AAA..AA}\to\color{red}\texttt{B}\color{black}\texttt{AAA..AAB}\),跳过大段 \(\texttt{A}\) 即有状态 \(\color{grey}\texttt{BAAA..AA}\color{red}\texttt{B}\)\(\color{green}\texttt{AAAA}\color{black}\xrightarrow{1}\color{green}\texttt{B}\color{black}\ (\text{add B})\)
  • \(\color{red}\texttt{A}\color{black}\texttt{BBB..BB}\to\color{red}\texttt{A}\color{black}\texttt{BBB..BBA}\),跳过大段 \(\texttt{B}\) 即有状态 \(\color{grey}\texttt{ABBB..BB}\color{red}\texttt{A}\)\(\color{green}\texttt{BBBB}\color{black}\xrightarrow{1}\color{green}\texttt{A}\color{black}\ (\text{add A})\)
  • \(\color{red}\texttt{A}\color{black}\texttt{BBB..BB}\to\color{red}\texttt{A}\color{black}\texttt{BBB..BBB}\),不用动即为原状态,\(\color{green}\texttt{BBBB}\color{black}\xrightarrow{0}\color{green}\texttt{BBBB}\color{black}\ (\text{add B})\)

对于状态 \(\color{green}\texttt{AB}\)\(\color{green}\texttt{BA}\),我们有:

  • \(\texttt{?}\color{red}\texttt{A}\color{black}\texttt{B}\to\texttt{?}\color{red}\texttt{A}\color{black}\texttt{BA}\),发现不能不走,而走一步可以到 \(\color{grey}\texttt{?A}\color{red}\texttt{B}\color{grey}\texttt{A}\)(往后一步)或 \(\color{grey}\texttt{?AB}\color{red}\texttt{A}\)(跳过 \(\texttt{B}\)),此时同时可得两个状态 \(\color{green}\texttt{A}\)\(\color{green}\texttt{B}\),故 \(\color{green}\texttt{AB}\color{black}\xrightarrow{1}\color{green}\texttt{@}\color{black}\ (\text{add A})\)
  • \(\texttt{?}\color{red}\texttt{A}\color{black}\texttt{B}\to\texttt{?}\color{red}\texttt{A}\color{black}\texttt{BB}\),很明显,无论后续再加 \(\texttt{A/B}\) 都会跳过中间的 \(\texttt{B}\),故 \(\color{green}\texttt{AB}\color{black}\xrightarrow{0}\color{green}\texttt{BBBB}\color{black}\ (\text{add B})\)
  • \(\texttt{?}\color{red}\texttt{B}\color{black}\texttt{A}\to\texttt{?}\color{red}\texttt{B}\color{black}\texttt{AA}\),同上,\(\color{green}\texttt{BA}\color{black}\xrightarrow{0}\color{green}\texttt{AAAA}\color{black}\ (\text{add A})\)
  • \(\texttt{?}\color{red}\texttt{B}\color{black}\texttt{A}\to\texttt{?}\color{red}\texttt{B}\color{black}\texttt{AB}\),同上上上,\(\color{green}\texttt{BA}\color{black}\xrightarrow{1}\color{green}\texttt{@}\color{black}\ (\text{add B})\)

对于状态 \(\color{green}\texttt{@}\),我们有:

  • \(\color{red}\texttt{?}\color{black}\to\color{red}\texttt{?}\color{black}\texttt{A}\),不用走即有状态 \(\color{red}\texttt{?}\color{black}\texttt{A}\)\(\color{green}\texttt{@}\color{black}\xrightarrow{0}\color{green}\texttt{@A}\color{black}\ (\text{add A})\)
  • \(\color{red}\texttt{?}\color{black}\to\color{red}\texttt{?}\color{black}\texttt{B}\),不用走即有状态 \(\color{red}\texttt{?}\color{black}\texttt{B}\)\(\color{green}\texttt{@}\color{black}\xrightarrow{0}\color{green}\texttt{@A}\color{black}\ (\text{add B})\)

对于状态 \(\color{green}\texttt{@A}\)\(\color{green}\texttt{@B}\),我们有:

  • \(\color{red}\texttt{?}\color{black}\texttt{A}\color{black}\to\color{red}\texttt{?}\color{black}\texttt{AA}\),最优显然是将 \(\texttt{?}\to\texttt{B}\) 然后跳过大段 \(\texttt{A}\),所以 \(\color{green}\texttt{@A}\color{black}\xrightarrow{0}\color{green}\texttt{AAAA}\color{black}\ (\text{add A})\)
  • \(\color{red}\texttt{?}\color{black}\texttt{A}\color{black}\to\color{red}\texttt{?}\color{black}\texttt{AB}\),走一步可以到 \(\color{grey}\texttt{A}\color{red}\texttt{A}\color{grey}\texttt{B}\)(将 \(\texttt{?}\to\texttt{A}\) 然后向后走一步)或 \(\color{grey}\texttt{BA}\color{red}\texttt{B}\)(将 \(\texttt{?}\to\texttt{B}\) 然后向后走跳一步),此时同时可得两个状态 \(\color{green}\texttt{A}\)\(\color{green}\texttt{B}\),所以 \(\color{green}\texttt{@A}\color{black}\xrightarrow{1}\color{green}\texttt{@}\color{black}\ (\text{add B})\)
  • \(\color{red}\texttt{?}\color{black}\texttt{B}\color{black}\to\color{red}\texttt{?}\color{black}\texttt{BA}\),同上,\(\color{green}\texttt{@B}\color{black}\xrightarrow{1}\color{green}\texttt{@}\color{black}\ (\text{add A})\)
  • \(\color{red}\texttt{?}\color{black}\texttt{B}\color{black}\to\color{red}\texttt{?}\color{black}\texttt{BB}\),同上上上,\(\color{green}\texttt{@B}\color{black}\xrightarrow{1}\color{green}\texttt{@}\color{black}\ (\text{add B})\)

走出厕所

最后定义 \(dp_{i,j,s}\) 表示前 \(i\) 个点走了 \(j\) 步,当前所在状态为 \(s\),方案数。

初始化 \(dp_{0,0,\color{green}\texttt{?}\color{black}}=1\),跑上述自动机刷表转移即可。

小细节:状态所在的位置不是最终 \(n+1\) 点所在位置,所以还需要记一下 \(dis_s\) 表示状态 \(s\) 还要走多少步到达终点。

具体地,\(dis_{\color{green}\texttt{AA}}=dis_{\color{green}\texttt{BB}}=dis_{\color{green}\texttt{AAA}}=dis_{\color{green}\texttt{BBB}}=2\),其余均为 \(1\)。判一下统计答案即可。

实现

可以用 map 记录每个状态的 id,这样写转移会清晰很多。

code,时间复杂度 \(O(nk|S|)\),其中 \(|S|=13\)

posted @ 2024-05-31 22:48  liangbowen  阅读(46)  评论(0编辑  收藏  举报