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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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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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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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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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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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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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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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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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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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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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neo4j︱Cypher 查询语言简单案例（二）

其中节点之间用:’–>’，包含关系的可以用’->’ 1.2 关系有以下几种关系型写法： -[rel:KNOWS]-> //单关系，关系的变量名为rel，类型是 KNOWS (a)-[:KNOWS|:LIKE...]->(b) //查询a到b的两类关系 -[{since:2010}]-> //附加属性 -[:KNOWS*..4]-> example: a -[rel: type*minHop..maxHop]->...friend-of-a-friend (user)-[:KNOWS]-(friend)-[:KNOWS]-(foaf) shortest path: path = shortestPath( (user...)-[:KNOWS*..5]-(other) ) collaborative filtering (user)-[:PURCHASED]->(product)<-[:PURCHASED]-()-[:

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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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python import 原理剖析

实际 module 查找顺序 Python’s default sys.meta_path has three meta path finders, one that knows how to import...built-in modules, one that knows how to import frozen modules, and one that knows how to import modules...除了第二章说的3个路径，实际前置还有一层 cache sys.modules，The module cache. sys.meta_path: 一般有3个finders one that knows how...to import built-in modules one that knows how to import frozen modules one that knows how to import...how to import built-in modules, one that knows how to import frozen modules, and one that knows how

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HDUOJ---1213How Many Tables

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Cypher查询语句

MATCH (a:Person {name: 'Alice'}), (b:Person {name: 'Bob'})CREATE (a)-[:KNOWS]->(b)查询MATCH：用于匹配图形中的模式。...MATCH (n:Person)WITH nMATCH (n)-[:KNOWS]->(friends)RETURN n.name, COUNT(friends) AS friendsCount模式匹配Cypher...*可以指定关系的长度，如KNOWS*2表示匹配长度为 2 的关系。...MERGE (a:Person {name: 'Alice'})-[:KNOWS]->(b:Person {name: 'Bob'})删除操作DELETE：用于删除节点或关系。...MATCH (n:Person {name: 'Alice'})-[r:KNOWS]->()DELETE r更新操作SET：用于更新节点或关系的属性。

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HDU 1213 How Many Tables

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CodeForces 670C Cinema(排序，离散化)

Each of the scientists knows exactly one language....The scientist, who came to the movie, will be very pleased if he knows the audio language of the movie..., will be almost satisfied if he knows the language of subtitles and will be not satisfied if he does...integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the index of a language, which the i-th scientist knows

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点击加载更多

18:Tomorrow never knows？

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

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

Array - 277 Find the Celebrity

知识图谱（2）——neo4j的用法

K6 在 Nebula Graph 上的压测实践

图数据库neo4j(二)python 连接neo4j

Neo4j学习（3）：操作图数据库的语言--Cypher

neo4j︱与python结合的py2neo使用教程（四）

图数据库｜Nebula Graph v3.1.0 性能报告

解锁 Leetcode 新题：寻找明星

LeetCode 277. 搜寻名人（思维题）

众里寻他千百度：找网红算法

neo4j︱Cypher 查询语言简单案例（二）

HDU1213 （并查集）

python import 原理剖析

HDUOJ---1213How Many Tables

Cypher查询语句

HDU 1213 How Many Tables

CodeForces 670C Cinema(排序，离散化)

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