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敏感词和谷歌搜索连接重置
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敏感词和谷歌搜索连接重置# Joke - 肚皮舞运动
j*y
1
想用碎芝麻包花卷,但挺难搞碎,有什么机器推荐吗?
家里有blender,不知道能捣碎吗?
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b*y
2
A graph is given. You need to design a data structure with minimum space
complexity such that it does the follows
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k*i
3
两子弟在长安街斗殴,一子弟曰:你等着,我爹是敏感词。另一子弟听后一笑:你丫牛逼个屁,你爹敏感词,我爹在Google搜索连接被重置。
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s*n
4
grinder
还有简单的,直接买黑芝麻粉

【在 j******y 的大作中提到】
: 想用碎芝麻包花卷,但挺难搞碎,有什么机器推荐吗?
: 家里有blender,不知道能捣碎吗?

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c*2
5
"Finds whether nodes u and v have a path in between them in O(1) time."
Change path to adjacency or path with length 1? Otherwise, not quite
possible to do that in O(1)
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H*7
6
不懂
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i*n
7
建议直接上芝麻粉
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b*y
8
是careercup 上面amazon的一道题
http://www.careercup.com/question?id=3214677
Finds whether nodes u and v have a path in between them in O(1) time
用adjacency matrix可以实现;
Finds whether there is a path of length k between u and v in O(k)
这个不是很明白。
careercup上面也没讨论清楚。
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c*o
9
没看懂。

牛逼个屁,你爹敏感词,我爹在Google搜索连接被重置。

【在 k**********i 的大作中提到】
: 两子弟在长安街斗殴,一子弟曰:你等着,我爹是敏感词。另一子弟听后一笑:你丫牛逼个屁,你爹敏感词,我爹在Google搜索连接被重置。
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j*y
10
家里已经有很多芝麻了。
不能再进芝麻粉了。

【在 i****n 的大作中提到】
: 建议直接上芝麻粉
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b*y
11
ch222, I see what you mean...
normal adjacency matrix cannot find whether u,v have a path in O(1) time; it
can, if and only if the length of the path is 1 (i.e., adjecent).
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p*e
12
两个斑竹是搜索结果可能涉及不符合相关法律法规和政策的内容,未予显示。
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r*u
14
邻接矩阵里,每个元素存the lengths of all path from i to j. Then, search time
is O(k).

【在 b******y 的大作中提到】
: A graph is given. You need to design a data structure with minimum space
: complexity such that it does the follows

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d*e
16
所以要先预处理find the lengths of all paths?

time

【在 r**u 的大作中提到】
: 邻接矩阵里,每个元素存the lengths of all path from i to j. Then, search time
: is O(k).

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b*y
18
if it the "path" is the shortest path,
we can use adjacency matrix, each entry a[u][v] stores the length the
shortest path from u to v.
if u->v has no path, then set a[u][v] = -1
In this way, we can tell whether this is a path from u to v in O(1) by
checking if the entry is -1; we can also tell in O(1) whether there is a
shortest path of length k by checking whether the value of the entry is k.
if it the "path" is the simple path,
we can expand the adjacency matrix to be 3-D, i.e., a[u][v][i], the 3rd
dimension stores whether (1 or 0) there is a simple path of length i.
Therefore, the second operation can be done in O(1)
a[u][v][0] sets to 1 is there is no path. So, the first operation can also
be done in O(1).
But now the space complexity is O(n^3).
We can compress the 3rd dimension by using a linked list. In stead of using
a 3-D matrix, we can use a 2-D adjacency matrix, whose entry is a linked
list. A node of the linked list is i value (length of a simple path from u
to v, i.e., the i in a[u][v][i] previously). In the worst case, from u to v,
there are simple paths of length 1, 2, 3, ...,k, we can tell whether there
is a simple path of length k by traversing the linked list in O(k) time.
The space complexity is now O(k*n^2)
BTW, thanks to my advisor in China to guide me into this proposal!
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t*e
19
我是放在大碗里 然后用个小啤酒瓶底慢慢碾碎的。。

【在 j******y 的大作中提到】
: 想用碎芝麻包花卷,但挺难搞碎,有什么机器推荐吗?
: 家里有blender,不知道能捣碎吗?

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c*l
20
and: use adjacency matrix multiplication to populate the linked list in each
matrix entry
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l*s
21


【在 j******y 的大作中提到】
: 是把芝麻炒熟后再grind的吗?
: 谢谢!

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l*s
22
挺好用
但不知道寿命

【在 a******o 的大作中提到】
: 好东西呀,好用吗,心动
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x*t
23
put the sesame seeds on a baking sheet, and use an empty glass bottle (beer,
wine) to roll.
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P*D
24
一般的food processor就可以。我一直用food processor做芝麻糊的。
另外给你讲一下我的血泪教训:芝麻这种油料作物你就不要想磨成特别细的粉了,不然
磨出来的只能是小磨香油。我第一次做芝麻糊的时候用的是coffee grinder,结果出来
的是芝麻油泥,然后coffee grinder又很难彻底拆开,你不知道我费了多大力气才把那
破玩意儿洗干净……
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s*e
25
我是用的blender,从walmart买的10速的,里面的刀片特别快。先把芝麻炒熟后,冷却
后倒进blender,用6速转1分半钟就可以了,如果看着还有整的颗粒,就用筷子把里面
的芝麻搅拌一下,接着在打一会儿。

【在 j******y 的大作中提到】
: 想用碎芝麻包花卷,但挺难搞碎,有什么机器推荐吗?
: 家里有blender,不知道能捣碎吗?

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s*k
26

千万不能打时间长
会打出芝麻油的
我就是这么悲剧的 lol
btw
我都是用擀面杖碾 就放在塑料袋 然后像擀面一样赶来赶去 赶来赶去 当然了 要用
力 而且还要有把力气哈
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