Problem 1124 - Football Coach Description It is not an easy job to be a coach of a football team. The season is almost over, only a few matches are left to play. All of sudden the team manager comes to you and tells you bad news: the main sponsor of…

-----------------------------最优化问题------------------------------------- ----------------------常规动态规划  SOJ1162 I-Keyboard  SOJ1685 Chopsticks SOJ1679 Gangsters SOJ2096 Maximum Submatrix  SOJ2111 littleken bg SOJ2142 Cow Exhibition  SOJ2505 The County…

No matter how far you may fly, never forget where you come from. 无论你能飞多远,都别忘了你来自何方. I never forget where I come from, but obviously I haven't fly too far from the start point. Staying still means falling behind, so it seems most of my peers, even mos…

Barbie Exposure May Limit Girls' Career Imagination The ubiquitous Barbie doll: she's been everything from a football coach to a surgeon. But girls who play with Barbie may have their ambition stunted. That's according to a study in the journal Sex R…

3876: [Ahoi2014]支线剧情 Time Limit: 10 Sec  Memory Limit: 256 MBSubmit: 821  Solved: 502[Submit][Status][Discuss] Description [故事背景] 宅男JYY非常喜欢玩RPG游戏,比如仙剑,轩辕剑等等.不过JYY喜欢的并不是战斗场景,而是类似电视剧一般的充满恩怨情仇的剧情.这些游戏往往 都有很多的支线剧情,现在JYY想花费最少的时间看完所有的支线剧情. [问题描述] JYY现在所玩的R…

题目 Source https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3602 Description The administration of a well-known football team has made a study about the lack of support in international away games.…

时间限制:1000MS  内存限制:65535K 提交次数:0 通过次数:0 题型: 编程题   语言: C++;C Description As every one known, a football team has 11 players . Now, there is a big problem in front of the Coach Liu. The final contest is getting closer. Who is the center defense, the ful…

流的概念和作用 学习Java IO,不得不提到的就是JavaIO流. 流是一组有顺序的,有起点和终点的字节集合,是对数据传输的总称或抽象.即数据在两设备间的传输称为流,流的本质是数据传输,根据数据传输特性将流抽象为各种类,方便更直观的进行数据操作. IO流的分类 根据处理数据类型的不同分为:字符流和字节流 根据数据流向不同分为:输入流和输出流 字符流和字节流 字符流的由来: 因为数据编码的不同,而有了对字符进行高效操作的流对象.本质其实就是基于字节流读取时,去查了指定的码表.字节流和字符流的区别…

1: /** 2: ZOJ 3229 有上下界的最大流 3: 两次求最大流的过程,非二分 4: 有源汇上下界的最大流问题, 首先连接 sink -> src, [0,INF]. 5: 根据net的正负,来建立 Supersrc 与 supersink 之间的边,做一次 maxflow. 6: 若所有的Supersrc 与 Supersink满流,则说明存在可行流. 7: 然后删除 sink -> src之间的边.(cap 置零即可). 从src -> sink 做一次最大流. 8: 两次…

1: /** 2: POJ 3801 有上下界的最小流 3: 4: 1.对supersrc到supersink 求一次最大流,记为f1.(在有源汇的情况下,先使整个网络趋向必须边尽量满足的情况) 5: 2.添加一条边sink -> src,流量上限为INF,这条边记为p.(构造无源汇网络) 6: 3.对supersrc到supersink再次求最大流,记为f2,这里判断是否为可行流.(要判断可行,必须先构造无源汇网络流,因此要再次求最大流) 7: 8: 此网络流的最小流即为 sink -> s…