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Who Knows You Better

这一集有一个地方印象非常深刻,是警方和碎片的 COO 去了解 Chris 背景信息维度的差异。

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18:Tomorrow never knows

18:Tomorrow never knows? 总时间限制: 1000ms 内存限制: 65536kB描述 甲壳虫的《A day in the life》和《Tomorrow never knows》脍炙人口,如果告诉你a day in the life ,真的会是tomorrow never knows?

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    How does CRM middleware knows which sites are interested with CRM local changes

    So how does CRM middleware knows which external sites must be sent with BDOC?

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    How does CRM middleware knows which sites are interested with CRM local changes

    means the BDOC will be sent to multiple external sites besides ERP. [1240] So how does CRM middleware knows

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    Either the node already knows other nodes (check with CLUSTER NODES)

    Either the node already knows other nodes (check with CLUSTER NODES) or contains some key in database

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    Array - 277 Find the Celebrity

    to get information of whether A knows B. You are given a helper function bool knows(a, b) which tells you whether A knows B. knows 1,and 1 knows no one. 做这道题暴力解法brute force是可以做到的O(n^2)的时间复杂度,很明显需要优化,突破点就在于knows这个api。 每次调用knows(i,j)如果返回false,可以确定,j一定不是名人,返回true就可以确定这个i一定认识j,所以i一定不是名人,这样就一定会有一个人可能是名人。

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    NebulaGraph v3.3.0 性能报告,Match count QPS 有 2~8 倍提升,3 跳查询 QPS 时延低至原 13

    yield KNOWS.creationDate 一跳·吞吐率 图片 一跳·服务端耗时(ms) 图片 一跳·客户端耗时(ms) 一跳·请求返回行数 图片 二跳·吞吐率 图片 二跳·服务端耗时(ms) yield DISTINCT KNOWS.creationDate as t, $$.Person.firstName, $$.Person.lastName, $$.Person.birthday 查询带边属性 GO {} STEP FROM {} OVER KNOWS yield KNOWS.creationDate 一跳·吞吐率 图片 一跳·服务端耗时(ms) 一跳·客户端耗时(ms) 二跳· REVERSELY YIELD DISTINCT KNOWS. Go3StepDistinctDst GO 3 STEP FROM {0} OVER KNOWS REVERSELY YIELD DISTINCT KNOWS.

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    知识图谱(2)——neo4j的用法

    {since: 2001}]->(js),(ee)-[:KNOWS {rating: 5}]->(ir), (js)-[:KNOWS]->(ir),(js)-[:KNOWS]->(rvb), (ir) -[:KNOWS]->(js),(ir)-[:KNOWS]->(ally), (rvb)-[:KNOWS]->(ally) 创造关系,谁和谁认识,什么时候认识,什么年龄认识 (ee)-[:KNOWS { since: 2001}]->(js),(ee)-[:KNOWS {rating: 5}]->(ir) 用match定义knows就是frieds 返回friends查看emil的朋友是谁 MATCH 查看谁认识johan并且爱好是surfng MATCH (js:Person)-[:KNOWS]-()-[:KNOWS]-(surfer) WHERE js.name = "Johan" AND surfer.hobby 显示可视化查询方案 PROFILE MATCH (js:Person)-[:KNOWS]-()-[:KNOWS]-(surfer) WHERE js.name = "Johan" AND surfer.hobby

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    K6 在 Nebula Graph 上的压测实践

    2199023256684 OVER KNOWS,1377,2202,true,4, 1628147822,GO 1 STEP FROM 4398046515995 OVER KNOWS,1487,2017 933 OVER KNOWS,1130,3422,true,5, 1628147822,GO 1 STEP FROM 6597069771971 OVER KNOWS,1022,2292,true,60 OVER KNOWS,1252,1811,true,13, 1628147822,GO 1 STEP FROM 10995116279792 OVER KNOWS,1115,1858,true,6, ,GO 1 STEP FROM 13194139536109 OVER KNOWS,1027,1604,true,2, 1628147822,GO 1 STEP FROM 10027 OVER KNOWS ,GO 1 STEP FROM 17592186050570 OVER KNOWS,768,1630,true,26, 1628147822,GO 1 STEP FROM 8853 OVER KNOWS

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    图数据库neo4j(二)python 连接neo4j

    Relationship a = Node('Person', name='Alice') b = Node('Person', name='Bob') r = Relationship(a, 'KNOWS ', b) print(a, b, r) 输出结果: (:Person {name: 'Alice'}) (:Person {name: 'Bob'}) (Alice)-[:KNOWS {}]->(Bob 运行结果: (alice:Person {age:20,name:"Alice"}) (bob:Person {age:21,name:"Bob"}) (alice)-[:KNOWS {time:"2017 )-[:KNOWS]->(bob)-[:LIKES]->(mike)<-[:KNOWS]-(alice) 这样我们就形成了一个 Walkable 对象。 = RelatedTo('Person', 'KNOWS') person = Person.select(graph).where(age=21).first() print(list(person.knows

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    解锁 Leetcode 新题:寻找明星

    to get information of whether A knows B. You are given a helper function bool knows(a, b) which tells you whether A knows B. 回到题目本身,别忘了我们还有一个 helper function bool knows(a, b), 如何去灵活的运用这个 helper function? 让我们翻回头来思考一下名人的定义,再结合 knows() 函数所给出的结果: If Knows (a, b) == true, 说明 a 认识 b, 那么 a 是否还有可能成为 celebrity? If Knows (a, b) == false, 说明 a 不认识 b, 那么 b 可能是 celebrity么? 答案是否定的,因为根据定义,名人被所有人都认识,所以 b 不是名人。

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    neo4j︱与python结合的py2neo使用教程(四)

    Relationship a = Node("Person", name="Alice") b = Node("Person", name="Bob") ab = Relationship(a, "KNOWS ", b) >>> ab >>> (alice)-[:KNOWS]->(bob) 新建两个节点a、b,分别具有一个name属性值,还新建a与b之间有向关系ab,ab的label为KNOWS。 ", b) ac = Relationship(a, "KNOWS", c) w = ab + Relationship(b, "LIKES", c) + ac print(w) >>> (alice) -[:KNOWS]->(bob)-[:LIKES]->(mike)<-[:KNOWS]-(alice) 另外我们可以调用 walk() 方法实现遍历,实例如下: from py2neo import walk = RelatedTo('Person', 'KNOWS') person = Person.select(graph).where(age=21).first() print(list(person.knows

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    图数据库|Nebula Graph v3.1.0 性能报告

    Graph v3.1.0 性能测试报告] [Nebula Graph v3.1.0 性能测试报告] [Nebula Graph v3.1.0 性能测试报告] MatchTest4 MATCH (m)-[:KNOWS ]-(n) WHERE id(m)=={} OPTIONAL MATCH (n)<-[:KNOWS]-(l) RETURN length(m.Person.lastName) AS n1, length ]-(n) WHERE id(m)=={} MATCH (n)-[:KNOWS]-(l) WITH m AS x, n AS y, l RETURN x.Person.firstName AS n1, ]-(n) WHERE id(m)=={} OPTIONAL MATCH (n)<-[:KNOWS]-(l) RETURN length(m.Person.lastName) AS n1, length ]-(n) WHERE id(m)=={} MATCH (n)-[:KNOWS]-(l) WITH m AS x, n AS y, l RETURN x.Person.firstName AS n1,

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    LeetCode 277. 搜寻名人(思维题)

    在本题中,你可以使用辅助函数 bool knows(a, b) 获取到 A 是否认识 B。请你来实现一个函数 int findCelebrity(n)。 派对最多只会有一个 “名人” 参加。 解题 2.1 暴力解 /* The knows API is defined for you. bool knows(int a, int b); */ class Solution { public: int findCelebrity(int n) { int i, /j 认识 i count++; if(count < j+1)//有人不认识 i break;//不是名人 if(knows knows(i, ans))//认识别人或者有人不认识他 return -1; } return ans; } }; 164 ms 9.6 MB

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    Neo4j学习(3):操作图数据库的语言--Cypher

    4-1.png 这里创建了两个节点,并且Liu Da知道Chen Er,但Chen Er不知道Liu Da 查询这两个节点及关系 MATCH (p1: Person)-[KNOWS]-(p2: Person ) return p1,KNOWS,p2 ? MATCH (p1: Person)-[KNOWS]-(p2: Person) DELETE p1, KNOWS, p2 ? 5-1.png 查询所有节点 ? ]->(p3), (p3)-[:KNOWS]->(p2), (p3)-[:KNOWS]->(p4) ? 8-2.png MATCH (x) - [:KNOWS]- () - [:KNOWS] - (fighter) WHERE x.name = "Liu Er" and fighter.hobby = "

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    HDU1213 (并查集)

    One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and

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    Product settype在CRM WebClient UI架构中的地位

    Product settype acts as a very important role in CRM WebClient UI architecture. [1240] The GenIL layer knows Instead, it only knows product BOL model attributes. The API knows nothing about attributes modeled in Genil model. Instead, it only knows attributes modeled in settype. [1240] In GenIL implementation, there is a util

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    众里寻他千百度:找网红算法

    刚才也说了,knows函数底层就是在访问一个二维的邻接矩阵,一次调用的时间复杂度是 O(1),所以这个暴力解法整体的最坏时间复杂度是 O(N^2)。 那么,是否有其他高明的办法来优化时间复杂度呢? 综上,只要观察任意两个之间的关系,就至少能确定一个人不是名人,上述情况判断可以用如下代码表示: if (knows(cand, other) || ! knows(other, cand) || knows(cand, other)) { return -1; } } // cand 是名人 knows(other, cand) || knows(cand, other)) { // cand 不可能是名人,排除 // 假设 other 是名人 knows(other, cand) || knows(cand, other)) { return -1; } } return cand;

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    HDUOJ---1213How Many Tables

    One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and

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    Product settype acts as a very important role in CRM WebClient UI architecture

    The GenIL layer knows nothing about attributes modeled in settype. Instead, it only knows product BOL model attributes. The API knows nothing about attributes modeled in Genil model. Instead, it only knows attributes modeled in settype. ?

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