UVa 1627 Team them up! 题目: Team them up! Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description   Your task is to divide a number of persons into two teams, in such a way, that: everyone belongs to one of…

Your task is to divide a number of persons into two teams, in such a way, that: everyone belongs to one of the teams; every team has at least one member; every person in the team knows every other person in his team; teams are as close in their sizes…

https://cn.vjudge.net/problem/UVA-1627 题目 有n(n≤100)个人,把他们分成非空的两组,使得每个人都被分到一组,且同组中的人相互认识.要求两组的成员人数尽量接近.多解时输出任意方案,无解时输出No Solution. 例如,1认识2, 3, 5:2认识1, 3, 4, 5:3认识1, 2, 5,4认识1, 2, 3,5认识1, 2, 3, 4(注意4认识1但1不认识4),则可以分两组:{1,3,5}和{2,4}. 题解 不是互相认识的连边,然后保证一条边…

题意:给n个分成两个组,保证每个组的人都相互认识,并且两组人数相差最少,给出一种方案. 析:首先我们可以知道如果某两个人不认识,那么他们肯定在不同的分组中,所以我们可以根据这个结论构造成一个图,如果两个不相互认识, 那么就加一条边,然后如果这个图是二分图,那么这分组是可以,否则就是不可能的.然后dp[i][j]表示那两个组相差人数为 j 是不是可以达到, 当然可能为负数,所以可以提前加上n,然后就是逆序输出答案即可. 代码如下: #pragma comment(linker, "/STACK:1…

UVA.540 Team Queue (队列) 题意分析 有t个团队正在排队,每次来一个新人的时候,他可以插入到他最后一个队友的身后,如果没有他的队友,那么他只能插入到队伍的最后.题目中包含以下操作: 1.ENQUEUE x :表示编号为x的入队: 2.DEQUEUE:长队的队首出队. 3.STOP:停止模拟 并且对于每一个DEQUEUE操作,输出队首的编号. 如果我们直接用一个队列来模拟的话,是无法实现的,原因在于,我们无法向队列中间插入元素.那么题目中还有一条重要的性质,那么就是:可以插入到…

[Link]: [Description] 给你n个人; 有一些人之间有认识关系 a认识b,b不一定认识a 让你把这n个人分成两组 使得这两组中的每一组: 组内的人与人之间都相互认识. 并且,使得两组的人数之差尽可能小; 输出分组的方案; [Solution] 如果A和B不是相互认识 那么他们俩肯定是在不同的组 则,在不是相互认识的人之间建一条边; 这样就能形成一张图了; 并且可能有多个连通块 然后,对每个连通块做二分染色; 如果全都能做二分染色; 则有解 这样,每个连通块都有的点被染成0,有的…

首发:https://mp.csdn.net/mdeditor/80294426 例题5-6 团体队列(Team Queue,UVa540) 有t个团队的人正在排一个长队.每次新来一个人时,如果他有队友在排队,那么这个 新人会插队到最后一个队友的身后.如果没有任何一个队友排队,则他会排到长队的队尾. 输入每个团队中所有队员的编号,要求支持如下3种指令(前两种指令可以穿插进 行). ENQUEUEx:编号为x的人进入长队. DEQUEUE:长队的队首出队. STOP:停止模拟. 对于每个DEQUE…

题目代号:UVA 540 题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=481 题目描述: Team Queue Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2645 Ac…

题意:给出t个团体,这t个团体排在一起,每次新来一个x排队,如果在整个的团体队列中,有x的队友,那么x排在它的队友的后面,如果他没有队友,则排在长队的队尾 求给出的每一个出队命令,输出出队的人的编号 紫书上的思路:有两个队列,一个是每一个团体内部形成的队列,还有一个是这t个团体又构成的队列 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<st…

思路:使用优先队列,按队伍出现的时刻和自身出现的时刻定义优先级,同时记录此时刻队列里是否有自己队伍的人,一开始没注意,wa了两发. #include<map> #include<queue> #include<cstdio> #include<string> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const…

又是一道比较复杂的模拟题.题中有两种队列,一种是总队列,从前向后.其他的是各个团体的小队列,因为入队的人如果有队友的话,会优先进入团体队列. 所以我们先设置两个队列和一个map,设置map倒是可以不用担心开多大数组这样的问题.然后开两个队列,一个是基本的q1,另一个是q2[1010] 一般的STL容器如string vector deque都是有两种直接初始化的方法(不严谨)一种是a(maxn)这个和一维数组差不多,直接赋给里面多少元素,另一种就是a[maxn]这个差不多是一个容器的数组,每一个…

题意: 每个人都属于一个团体,在排队的时候,如果他所在的团体有人在队伍中,则他会站到这个团体的最后.否则站到整个队伍的队尾. 输出每次出队的人的编号. 分析: 容易看出,长队中,在同一个团体的人是排在一起的. 所以用两个队列模拟即可,一个队列保留团体的编号,另外一个队列数组存放的是团体中每个人的编号. #include <cstdio> #include <queue> #include <map> using namespace std; + ; map<int…

版权声明:本文为博主原创文章.未经博主同意不得转载. https://blog.csdn.net/u013840081/article/details/26180081 题目例如以下: Team Queue  Queues and Priority Queues are data structures which are known to most computer scientists. TheTeam Queue, however, is not so well known, though…

Team Queue Descriptions: Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 用两个队列模拟就好. 记录某个队在不在队列里面. 模拟 [错的次数] 在这里输入错的次数 [反思] 在这里输入反思 [代码] #include <bits/stdc++.h> using namespace std; const int N = 1e3; int t; queue <int> team[N+10]; queue <int> dl; map <int,int> dic; boo…

  Queues and Priority Queues are data structures which are known to most computer scientists. The Team Queue, however, is not so well known, though it occurs often in everyday life. At lunch time the queue in front of the Mensa is a team queue, for e…

uva 6757 Cup of CowardsCup of Cowards (CoC) is a role playing game that has 5 different characters (Mage, Tank, Fighter,Assassin and Marksman). A team consists of 5 players (one from each kind) and the goal is to kill amonster with L life points. The…

链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=1006 Problem A Useless Tile Packers Input: standard input Output: standard output Yes, as you have apprehended the Useless Tile Pac…

UVA - 10129Play on Words Some of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solveit to open that doors. Because there is no other way to open the doors, the puzzle is very importantfor us.There is a lar…

Description  UVa Panel Discussion  The UVa online judge team is arranging a panel discussion for the next ACM-ICPC World Finals event in Orlando, Florida. They want that three or four of the contestants take part in the panel and as they have about 3…

UVA11039-Building designing Time limit: 3.000 seconds An architect wants to design a very high building. The building will consist of some floors, and each floor has a certain size. The size of a floor must be greater than the size of the floor immediate…

:W Problem A: Football (aka Soccer)  The Problem Football the most popular sport in the world (americans insist to call it "Soccer",but we will call it "Football"). As everyone knows, Brasil is the country that have mostWorld Cup title…

关于欧拉回路和欧拉路径 定义:欧拉回路:每条边恰好只走一次,并能回到出发点的路径欧拉路径:经过每一条边一次,但是不要求回到起始点 ①首先看欧拉回路存在性的判定: 一.无向图每个顶点的度数都是偶数,则存在欧拉回路. 二.有向图(所有边都是单向的)每个节顶点的入度都等于出度,则存在欧拉回路. ②.欧拉路径存在性的判定 一.无向图一个无向图存在欧拉路径,当且仅当   该图所有顶点的度数为偶数   或者  除了两个度数为奇数外其余的全是偶数. 二.有向图一个有向图存在欧拉路径,当且仅当  该图所有顶点的…

SOLUTION VERIFIED September 13 2016 KB1248793 Environment Red Hat Enterprise Linux 7 NetworkManager teamd Issue Would like to configure a VLANs with Teaming in RHEL 7. Need to know how to make VLANs like team0.10 for a team0 device with RHEL and Netw…

SOLUTION VERIFIED September 13 2016 KB2620131 Environment Red Hat Enterprise Linux 7 NetworkManager teamd Issue Would like to configure a basic Team in RHEL 7. Need to know how to make team0 with RHEL. Resolution For steps to use Teaming with VLANs,…

h3{ color: #000; padding: 5px; margin-bottom: 10px; font-weight: bolder; background-color: #ccc; } h4 { color: #000; border-bottom: dashed 1px #ccc; padding-bottom: 5px; margin-bottom: 10px; font-weight: bolder; } 为什么软件项目需要 Team Leader 多年以前,当我接触敏捷时,我…

SOLUTION IN PROGRESS February 29 2016 KB2181361 environment Red Hat Enterprise Linux 7 Teaming,Bridge NetworkManager-1.0.6-27 and above question Is it possible to add a bridge interface over a team interface? ens3 --| |--- team0 -- bridge0 + (ipv4 ad…

4742: [Usaco2016 Dec]Team Building Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 21  Solved: 16[Submit][Status][Discuss] Description Every year, Farmer John brings his NN cows to compete for "best in show" at the state fair. His arch -rival, F…

矩阵式管理,是常见的经典管理架构.其最早起源于美国的航空航天部门,然后被美国人带到了日本,然后被日本人带到了台湾,然后台湾人带到大陆...矩阵管理最典型的特征是,组织架构按职能与专业划分,项目由跨越部门的人员组成.典型缺点是,由于项目负责人没有人事管理权,导致其责任与权力不匹配.当然,这样的缺点并没有影响其江湖地位,相反,它广受赞誉. 同时,有另外一些人,他们坚持“特性团队”(feature team)用有更高的效率. 1. 什么是 feature team 其实我们对 feature team…

自2014年9月起,安天AVL移动安全团队持续检测到一类基于Android移动平台的间谍类病毒,病毒样本大多伪装成名为"最高人民检察院"的应用.经过反编译逆向分析以及长期的跟踪调查,我们判断这可能是一起有组织的电信诈骗犯罪活动. 2014年9月至今,某诈骗组织持续以涉嫌犯罪为由恐吓受害者,并进行电信诈骗活动.整个诈骗流程大致如下:攻击者首先向受害者的手机发送含恶意应用下载链接的短信:攻击者通常以获取"案件号"."单位代号"."电子凭证&…