Degree Sequence of Graph G

Problem Description
Wang Haiyang is a strong and optimistic Chinese youngster. Although born and brought up in the northern inland city Harbin, he has deep love and yearns for the boundless oceans. After graduation, he came to a coastal city and got a job in a marine transportation company. There, he held a position as a navigator in a freighter and began his new life.

The cargo vessel, Wang Haiyang worked on, sails among 6 ports between which exist 9 routes. At the first sight of his navigation chart, the 6 ports and 9 routes on it reminded him of Graph Theory that he studied in class at university. In the way that Leonhard Euler solved The Seven Bridges of Knoigsberg, Wang Haiyang regarded the navigation chart as a graph of Graph Theory. He considered the 6 ports as 6 nodes and 9 routes as 9 edges of the graph. The graph is illustrated as below.

According to Graph Theory, the number of edges related to a node is defined as Degree number of this node.

Wang Haiyang looked at the graph and thought, If arranged, the Degree numbers of all nodes of graph G can form such a sequence: 4, 4, 3,3,2,2, which is called the degree sequence of the graph. Of course, the degree sequence of any simple graph (according to Graph Theory, a graph without any parallel edge or ring is a simple graph) is a non-negative integer sequence?

Wang Haiyang is a thoughtful person and tends to think deeply over any scientific problem that grabs his interest. So as usual, he also gave this problem further thought, As we know, any a simple graph always corresponds with a non-negative integer sequence. But whether a non-negative integer sequence always corresponds with the degree sequence of a simple graph? That is, if given a non-negative integer sequence, are we sure that we can draw a simple graph according to it.?

Let's put forward such a definition: provided that a non-negative integer sequence is the degree sequence of a graph without any parallel edge or ring, that is, a simple graph, the sequence is draw-possible, otherwise, non-draw-possible. Now the problem faced with Wang Haiyang is how to test whether a non-negative integer sequence is draw-possible or not. Since Wang Haiyang hasn't studied Algorithm Design course, it is difficult for him to solve such a problem. Can you help him?

Input
The first line of input contains an integer T, indicates the number of test cases. In each case, there are n+1 numbers; first is an integer n (n<1000), which indicates there are n integers in the sequence; then follow n integers, which indicate the numbers of the degree sequence.

Output
For each case, the answer should be "yes"or "no" indicating this case is "draw-possible" or "non-draw-possible"

Sample Input
2
6 4 4 3 3 2 2
4 2 1 1 1

Sample Output
yes
no

Problem Description Wang Haiyang is a strong and optimistic Chinese youngster. Although born and brought up in the northern inland city Harbin, he has deep love and yearns for the boundless oceans. After graduation, he came to a coastal city and got a job in a marine transportation company. There, he held a position as a navigator in a freighter and began his new life. The cargo vessel, Wang Haiyang worked on, sails among 6 ports between which exist 9 routes. At the first sight of his navigation chart, the 6 ports and 9 routes on it reminded him of Graph Theory that he studied in class at university. In the way that Leonhard Euler solved The Seven Bridges of Knoigsberg, Wang Haiyang regarded the navigation chart as a graph of Graph Theory. He considered the 6 ports as 6 nodes and 9 routes as 9 edges of the graph. The graph is illustrated as below. According to Graph Theory, the number of edges related to a node is defined as Degree number of this node. Wang Haiyang looked at the graph and thought, If arranged, the Degree numbers of all nodes of graph G can form such a sequence: 4, 4, 3,3,2,2, which is called the degree sequence of the graph. Of course, the degree sequence of any simple graph (according to Graph Theory, a graph without any parallel edge or ring is a simple graph) is a non-negative integer sequence? Wang Haiyang is a thoughtful person and tends to think deeply over any scientific problem that grabs his interest. So as usual, he also gave this problem further thought, As we know, any a simple graph always corresponds with a non-negative integer sequence. But whether a non-negative integer sequence always corresponds with the degree sequence of a simple graph? That is, if given a non-negative integer sequence, are we sure that we can draw a simple graph according to it.? Let's put forward such a definition: provided that a non-negative integer sequence is the degree sequence of a graph without any parallel edge or ring, that is, a simple graph, the sequence is draw-possible, otherwise, non-draw-possible. Now the problem faced with Wang Haiyang is how to test whether a non-negative integer sequence is draw-possible or not. Since Wang Haiyang hasn't studied Algorithm Design course, it is difficult for him to solve such a problem. Can you help him? Input The first line of input contains an integer T, indicates the number of test cases. In each case, there are n+1 numbers; first is an integer n (n<1000), which indicates there are n integers in the sequence; then follow n integers, which indicate the numbers of the degree sequence. Output For each case, the answer should be "yes"or "no" indicating this case is "draw-possible" or "non-draw-possible" Sample Input 2 6 4 4 3 3 2 2 4 2 1 1 1 Sample Output yes no
K-Anonymous Sequence
Description The explosively increasing network data in various application domains has raised privacy concerns for the individuals involved. Recent studies show that simply removing the identities of nodes before publishing the graph/social network data does not guarantee privacy. The structure of the graph itself, along with its basic form the degree of nodes, can reveal the identities of individuals. To address this issue, we study a specific graph-anonymization problem. We call a graph k-anonymous if for every node v, there exist at least k-1 other nodes in the graph with the same degree as v. And we are interested in achieving k-anonymous on a graph with the minimum number of graph-modification operations. We simplify the problem. Pick n nodes out of the entire graph G and list their degrees in ascending order. We define a sequence k-anonymous if for every element s, there exist at least k-1 other elements in the sequence equal to s. To let the given sequence k-anonymous, you could do one operation only—decrease some of the numbers in the sequence. And we define the cost of the modification the sum of the difference of all numbers you modified. e.g. sequence 2, 2, 3, 4, 4, 5, 5, with k=3, can be modified to 2, 2, 2, 4, 4, 4, 4, which satisfy 3-anonymous property and the cost of the modification will be |3-2| + |5-4| + |5-4| = 3. Give a sequence with n numbers in ascending order and k, we want to know the modification with minimal cost among all modifications which adjust the sequence k-anonymous. Input The first line of the input file contains a single integer T (1 ≤ T ≤ 20) – the number of tests in the input file. Each test starts with a line containing two numbers n (2 ≤ n ≤ 500000) – the amount of numbers in the sequence and k (2 ≤ k ≤ n). It is followed by a line with n integer numbers—the degree sequence in ascending order. And every number s in the sequence is in the range [0, 500000]. Output For each test, output one line containing a single integer—the minimal cost. Sample Input 2 7 3 2 2 3 4 4 5 5 6 2 0 3 3 4 8 9 Sample Output 3 5
Graph Reconstruction
Let there be a simple graph with N vertices but we just know the degree of each vertex. Is it possible to reconstruct the graph only by these information? A simple graph is an undirected graph that has no loops (edges connected at both ends to the same vertex) and no more than one edge between any two different vertices. The degree of a vertex is the number of edges that connect to it. Input There are multiple cases. Each case contains two lines. The first line contains one integer N (2 ≤ N ≤ 100), the number of vertices in the graph. The second line conrains N integers in which the ith item is the degree of ith vertex and each degree is between 0 and N-1(inclusive). Output If the graph can be uniquely determined by the vertex degree information, output "UNIQUE" in the first line. Then output the graph. If there are two or more different graphs can induce the same degree for all vertices, output "MULTIPLE" in the first line. Then output two different graphs in the following lines to proof. If the vertex degree sequence cannot deduced any graph, just output "IMPOSSIBLE". The output format of graph is as follows: N E u1 u2 ... uE v1 v2 ... vE Where N is the number of vertices and E is the number of edges, and {ui,vi} is the ith edge the the graph. The order of edges and the order of vertices in the edge representation is not important since we would use special judge to verify your answer. The number of each vertex is labeled from 1 to N. See sample output for more detail. Sample Input 1 0 6 5 5 5 4 4 3 6 5 4 4 4 4 3 6 3 4 3 1 2 0 Sample Output UNIQUE 1 0 UNIQUE 6 13 3 3 3 3 3 2 2 2 2 1 1 1 5 2 1 5 4 6 1 5 4 6 5 4 6 4 MULTIPLE 6 12 1 1 1 1 1 5 5 5 6 6 2 2 5 4 3 2 6 4 3 2 4 3 4 3 6 12 1 1 1 1 1 5 5 5 6 6 3 3 5 4 3 2 6 4 3 2 4 2 4 2 IMPOSSIBLE
Strange Graph
Description Let us consider an undirected graph G = < V,E >. Let us denote by N(v) the set of vertices connected to vertex v (i.e. the set of neighbours of v). Recall that the number of vertices connected to v is called the degree of this vertex and is denoted by deg v. We will call graph G strange if it is connected and for its every vertex v the following conditions are satisfied: 1. deg v >= 2 (i.e. there are at least two vertices connected to v) 2. If deg v = 2 then the two neighbours of v are not connected by an edge 3. If degv > 2 then there is u ∈ N(v), such that the following is true: (a) deg u = 2 (b) Any two different vertices w1,w2 ∈ N(v) \ {u} are connected, i.e. (w1,w2) ∈ E. You are given some strange graph G. Find hamiltonian cycle in it, i.e. find such cycle that it goes through every vertex of G exactly once. Input The first line of the input file contains two integer numbers N and M -- the number of vertices and edges in G respectively (3 <= N <= 10 000, M <= 100 000). 2M integer numbers follow -- each pair represents vertices connected by the corresponding edge (vertices are numbered from 1 to N). It is guaranteed that each edge occurs exactly once in the input file and that there are no loops (i.e. ends of each edge are distinct). Output If there is no hamiltonian cycle in G, print -1 on the first line of the output file. In the other case output N numbers -- the sequence of vertices of G as they appear in the hamiltonian cycle found (note that the last vertex must be connected to the first one). If there are several solutions, output any one. Sample Input 4 4 1 2 2 3 3 4 4 1 Sample Output 1 2 3 4
Octagons 的问题
Problem Description Below is a picture of an infinite hyperbolic tessellation of octagons. If we think of this as a graph of vertices (of degree three), then there exists an isomorphism of the graph which maps any vertex x onto any other vertex y. Every edge is given a label from the set {a,b,c} in such a way that every vertex has all three types of edges incident on it, and the labels alternate around each octagon. Part of this labeling is illustrated in the diagram. So a path in this graph (starting from any vertex) can be specified by a sequence of edge labels. Your job is to write a program which, given a squence of labels such as "abcbcbcabcaccabb", returns "closed" if the path ends on the same vertex where it starts, and returns "open" otherwise. Input The input will begin with a number Z ≤ 200 on a line by itself. This is followed by Z lines, each of which is a squence of length at least 1 and at most 40 of 'a's 'b's and 'c's. Output For each input instance, the output will be the words "closed" or "open", each on a single line. Sample Input 2 abababab abcbcbcbcba Sample Output closed open
Octagons 的打开
Problem Description Below is a picture of an infinite hyperbolic tessellation of octagons. If we think of this as a graph of vertices (of degree three), then there exists an isomorphism of the graph which maps any vertex x onto any other vertex y. Every edge is given a label from the set {a,b,c} in such a way that every vertex has all three types of edges incident on it, and the labels alternate around each octagon. Part of this labeling is illustrated in the diagram. So a path in this graph (starting from any vertex) can be specified by a sequence of edge labels. Your job is to write a program which, given a squence of labels such as "abcbcbcabcaccabb", returns "closed" if the path ends on the same vertex where it starts, and returns "open" otherwise. Input The input will begin with a number Z ≤ 200 on a line by itself. This is followed by Z lines, each of which is a squence of length at least 1 and at most 40 of 'a's 'b's and 'c's. Output For each input instance, the output will be the words "closed" or "open", each on a single line. Sample Input 2 abababab abcbcbcbcba Sample Output closed open
The Postal Worker Rings Once
Description Graph algorithms form a very important part of computer science and have a lineage that goes back at least to Euler and the famous Seven Bridges of Konigsberg problem. Many optimization problems involve determining efficient methods for reasoning about graphs. This problem involves determining a route for a postal worker so that all mail is delivered while the postal worker walks a minimal distance, so as to rest weary legs. Given a sequence of streets (connecting given intersections) you are to write a program that determines the minimal cost tour that traverses every street at least once. The tour must begin and end at the same intersection. The ``real-life'' analogy concerns a postal worker who parks a truck at an intersection and then walks all streets on the postal delivery route (delivering mail) and returns to the truck to continue with the next route. The cost of traversing a street is a function of the length of the street (there is a cost associated with delivering mail to houses and with walking even if no delivery occurs). In this problem the number of streets that meet at a given intersection is called the degree of the intersection. There will be at most two intersections with odd degree. All other intersections will have even degree, i.e., an even number of streets meeting at that intersection. Input The input consists of a sequence of one or more postal routes. A route is composed of a sequence of street names (strings), one per line, and is terminated by the string ``deadend'' which is NOT part of the route. The first and last letters of each street name specify the two intersections for that street, the length of the street name indicates the cost of traversing the street. All street names will consist of lowercase alphabetic characters. For example, the name foo indicates a street with intersections f and o of length 3, and the name computer indicates a street with intersections c and r of length 8. No street name will have the same first and last letter and there will be at most one street directly connecting any two intersections. As specified, the number of intersections with odd degree in a postal route will be at most two. In each postal route there will be a path between all intersections, i.e., the intersections are connected. Output For each postal route the output should consist of the cost of the minimal tour that visits all streets at least once. The minimal tour costs should be output in the order corresponding to the input postal routes. Sample Input one two three deadend mit dartmouth linkoping tasmania york emory cornell duke kaunas hildesheim concord arkansas williams glasgow deadend Sample Output 11 114
The Postal Worker Rings Once
Description Graph algorithms form a very important part of computer science and have a lineage that goes back at least to Euler and the famous Seven Bridges of Konigsberg problem. Many optimization problems involve determining efficient methods for reasoning about graphs. This problem involves determining a route for a postal worker so that all mail is delivered while the postal worker walks a minimal distance, so as to rest weary legs. Given a sequence of streets (connecting given intersections) you are to write a program that determines the minimal cost tour that traverses every street at least once. The tour must begin and end at the same intersection. The ``real-life'' analogy concerns a postal worker who parks a truck at an intersection and then walks all streets on the postal delivery route (delivering mail) and returns to the truck to continue with the next route. The cost of traversing a street is a function of the length of the street (there is a cost associated with delivering mail to houses and with walking even if no delivery occurs). In this problem the number of streets that meet at a given intersection is called the degree of the intersection. There will be at most two intersections with odd degree. All other intersections will have even degree, i.e., an even number of streets meeting at that intersection. Input The input consists of a sequence of one or more postal routes. A route is composed of a sequence of street names (strings), one per line, and is terminated by the string ``deadend'' which is NOT part of the route. The first and last letters of each street name specify the two intersections for that street, the length of the street name indicates the cost of traversing the street. All street names will consist of lowercase alphabetic characters. For example, the name foo indicates a street with intersections f and o of length 3, and the name computer indicates a street with intersections c and r of length 8. No street name will have the same first and last letter and there will be at most one street directly connecting any two intersections. As specified, the number of intersections with odd degree in a postal route will be at most two. In each postal route there will be a path between all intersections, i.e., the intersections are connected. Output For each postal route the output should consist of the cost of the minimal tour that visits all streets at least once. The minimal tour costs should be output in the order corresponding to the input postal routes. Sample Input one two three deadend mit dartmouth linkoping tasmania york emory cornell duke kaunas hildesheim concord arkansas williams glasgow deadend Sample Output 11 114

《MySQL 性能优化》之理解 MySQL 体系结构

python自动下载图片

【前言】 　　收到一封来信，赶上各种事情拖了几日，利用今天要放下工作的时机，做个回复。 　　2020年到了，就以这一封信，作为开年标志吧。 【正文】 　　您好，我是一名现在有很多困惑的大二学生。有一些问题想要向您请教。 　　先说一下我的基本情况，高考失利，不想复读，来到广州一所大专读计算机应用技术专业。学校是偏艺术类的，计算机专业没有实验室更不用说工作室了。而且学校的学风也不好。但我很想在计算机领...

【CSDN编者按】1月2日，阿里巴巴发布《达摩院2020十大科技趋势》，十大科技趋势分别是：人工智能从感知智能向认知智能演进；计算存储一体化突破AI算力瓶颈；工业互联网的超融合；机器间大规模协作成为可能；模块化降低芯片设计门槛；规模化生产级区块链应用将走入大众；量子计算进入攻坚期；新材料推动半导体器件革新；保护数据隐私的AI技术将加速落地；云成为IT技术创新的中心 。 新的画卷，正在徐徐展开。...

Python+OpenCV实时图像处理

2020年一线城市程序员工资大调查

python爬取百部电影数据，我分析出了一个残酷的真相
2019年就这么匆匆过去了，就在前几天国家电影局发布了2019年中国电影市场数据，数据显示去年总票房为642.66亿元，同比增长5.4%；国产电影总票房411.75亿元，同比增长8.65%，市场占比 64.07%；城市院线观影人次17.27亿，同比增长0.64%。 看上去似乎是一片大好对不对？不过作为一名严谨求实的数据分析师，我从官方数据中看出了一点端倪：国产票房增幅都已经高达8.65%了，为什...

Windows可谓是大多数人的生产力工具，集娱乐办公于一体，虽然在程序员这个群体中都说苹果是信仰，但是大部分不都是从Windows过来的，而且现在依然有很多的程序员用Windows。 所以，今天我就把我私藏的Windows必装的软件分享给大家，如果有一个你没有用过甚至没有听过，那你就赚了????，这可都是提升你幸福感的高效率生产力工具哦！ 走起！???? NO、1 ScreenToGif 屏幕，摄像头和白板...

C++(数据结构与算法):62---搜索树（二叉搜索树、索引二叉搜索树）

AI 没让人类失业，搞 AI 的人先失业了

2020年，冯唐49岁：我给20、30岁IT职场年轻人的建议

B站是个宝，谁用谁知道???? 作为一名大学生，你必须掌握的一项能力就是自学能力，很多看起来很牛X的人，你可以了解下，人家私底下一定是花大量的时间自学的，你可能会说，我也想学习啊，可是嘞，该学习啥嘞，不怕告诉你，互联网时代，最不缺的就是学习资源，最宝贵的是啥？ 你可能会说是时间，不，不是时间，而是你的注意力，懂了吧！ 那么，你说学习资源多，我咋不知道，那今天我就告诉你一个你必须知道的学习的地方，人称...

leetcode88. 合并两个有序数组

Java校招入职华为，半年后我跑路了

Python全栈 Linux基础之3.Linux常用命令
Linux对文件（包括目录）有很多常用命令，可以加快开发效率：ls是列出当前目录下的文件列表，选项有-a、-l、-h，还可以使用通配符；c功能是跳转目录，可以使用相对路径和绝对路径；mkdir命令创建一个新的目录，有-p选项，rm删除文件或目录，有-f、-r选项；cp用于复制文件，有-i、-r选项，tree命令可以将目录结构显示出来（树状显示），有-d选项，mv用来移动文件/目录，有-i选项；cat查看文件内容，more分屏显示文件内容，grep搜索内容；>、>>将执行结果重定向到一个文件；|用于管道输出。
​两年前不知如何编写代码的我，现在是一名人工智能工程师

loonggg读完需要5分钟速读仅需 2 分钟大家好，我是你们的校长。我知道大家在家里都憋坏了，大家可能相对于封闭在家里“坐月子”，更希望能够早日上班。今天我带着大家换个思路来聊一个问题...
Spring框架|JdbcTemplate介绍