其他大平台, 如链家, 贝壳等往往不能短租; 但是可以参考他们提供的信息, 比如房价, 设施等, 根据这个制定 checklist.
在 Jekyll 版本的 NexT 主题上做了若干修改. 目前初步搬运完成, 还需要进一步 debug.
线下使用 VNote 写作, 目前发现的和线上 md 的区别:
[discription](url) 描述中如果有 | 的话线上会编译成 table, 需要转义.I have moved some contents (including cheatcode generators) to wiki to reduce clutter. (2020/11/1)
Here is the new post. (2021/1/2)
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