Tactics Ogre One Vision Mod
写了简易版, 若没有反馈就不再补充.
有问题可以直接问我.
Months ago I came across an article on the methods of shuffling cards in TCG tournaments. Given limited time in a match, the author, hyakkoTCG@hyakko_tcg sought a way to mix a deck thoroughly by shuffling three times. However, at least to me, it is obvious that the methodology proposed is fundamentally flawed. Later I found out in surprise that the article was widely disseminated: it got 2.3k retweets and 3.1k likes on twitter.
本系列起始于 2019 年夏天, 目的是记录自己学习厨艺的过程, 尽量把烹饪流程拆分成原子操作, 并探究其原理. 不过因为实践的次数极少所以进展龟速, 另外原理部分我是外行, 全靠查资料, 这进一步拖慢了进度. 姑且先发出来之后慢慢填坑.
2020/7/19
广度优先搜索.
一个 trick 是可以用二进制数表示访问过的节点, 1 表示访问过, 0 表示未访问过. 比如一共 5 个节点 (0-indexed), 则可以用 11001 表示访问过节点 0, 3, 4. 位运算 1<<n 比 2**n 快得多.
from collections import deque
class Solution:
def shortestPathLength(self, graph: List[List[int]]) -> int:
'''
binary representation
e.g. 11001 for nodes 034 visited and 12 unvisited
'''
goal = (1<<len(graph)) - 1
# (curr_node, visited_nodes, steps)
queue = deque((node, 1<<node, 0) for node in range(len(graph)))
seen = set()
while queue:
curr_node, visited_nodes, steps = queue.popleft()
if visited_nodes == goal:
return steps
for adj_node in graph[curr_node]:
state = (adj_node, visited_nodes | 1<<adj_node, steps+1)
if state not in seen:
seen.add(state)
queue.append(state)
return -1
要求 $a^n$, 其中 $a\in\mathbb R$, $n\in\mathbb Z$. 先不妨假设 $n\ge 0$, 基本想法是
\[a^n = \begin{cases} a^{n/2}a^{n/2}, & \text{if $n$ is even,}\\ a^{(n-1)/2}a^{(n-1)/2}a, & \text{if $n$ is odd.} \end{cases}\]很容易写出时间复杂度 $O(\log n)$ 的递归算法, 而要写迭代算法需要再想一想.
第二篇杂录 侧重最佳实践.
最近 (2021/10/28) 发现官方文档有 Programming FAQ — Python 3.10.0 documentation, 很有用.
2023/1/10
(1,) + (2, 3)
# (1, 2, 3)
a = {1: 1}
b = {2: 2}
{**a, **b}
# {1: 1, 2: 2}
2022/7/5
忘了看哪个源码的时候读到的
Python の nonlocal と global の違い - Qiita
lru_cache 以及 singledispatch