v-if和v-else指令求苹果总价

Vue.js 源码分析(十七) 指令篇 v-if、v-else-if和v-else 指令详解

v-if 指令用于条件性地渲染一块内容.这块内容只会在指令的表达式返回true值的时候被渲染. v-else-if,顾名思义,充当 v-if 的“else-if 块”,可以连续使用: 也可以使用 v-else 指令来表示 v-if 的“else 块”: 挺好理解的,就和大多数的语言的if()....else if()...else逻辑语句是一样的,例如: <!DOCTYPE html> <html lang="en"> <head> <meta

poj 2762 Going from u to v or from v to u?

题目描述:为了让他们的儿子变得更勇敢些,Jiajia和Wind将他们带到一个大洞穴中.洞穴中有n个房间,有一些单向的通道连接某些房间.每次,Wind选择两个房间x和y,要求他们的一个儿子从一个房间走到另一个房间,这个儿子可以从x走到y,也可以从y走到x.Wind保证她布置的任务是可以完成的,但她确实不知道如何判断一个任务是否可以完成.为了使Wind下达任务更容易些,Jiajia决定找这样的一个洞穴,每对房间(设为x和y)都是相通(可以从x走到y,或者可以从y走到x)的.给定一个洞穴,你能告诉Ji

poj 2762 Going from u to v or from v to u?(强连通分量+缩点重构图+拓扑排序)

http://poj.org/problem?id=2762 Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 12733 Accepted: 3286 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has

POJ 2762 Going from u to v or from v to u?（强联通 + TopSort）

题目大意: 为了锻炼自己的儿子 Jiajia 和Wind 把自己的儿子带入到一个洞穴内,洞穴有n个房间,洞穴的道路是单向的. 每一次Wind 选择两个房间 x 和 y, 让他的儿子从一个房间走到另一个房间,(要么从 x->y 或者 y->x), Wind承诺这个是一定可以走到的.但是他不知道如何判断这个 xy一定是互通的,现在给你一个洞穴,问随机给你两个洞穴的编号,是否是相通的. 题目分析:题目意思有点偏移,其实意思是(只要能从x->y 或者 y->x 这个都算是通的),

poj 2762 Going from u to v or from v to u? (推断它是否是一个薄弱环节图)

意甲冠军:给定一个有向图有m单向边缘.免费推断是否两点起来(a可以b要么b可以a或最多彼此),该请求弱联通重量. 算法: 缩点求强连通分量.然后又一次建图.推断新图是否是一条单链,即不能分叉,假设分叉了就会存在不可达的情况. 怎么推断是否是单链呢? 就是每次入度为0的点都仅仅有一个,即每次队列里仅仅有一个点. ( o(╯□╰)o.... .好像已经是第二次用pair记录原图的点对,然后存pair的vector忘记清空导致wa来wa去! ) #include<cstdio> #inclu

Going from u to v or from v to u? POJ - 2762（强连通有向最长路径）

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The so

poj 2762 Going from u to v or from v to u? trajan+拓扑

Going from u to v or from v to u? Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of th

[ tarjan + dfs ] poj 2762 Going from u to v or from v to u?

题目链接: http://poj.org/problem?id=2762 Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 14546 Accepted: 3837 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cav

POJ - 2762 Going from u to v or from v to u? (强连通缩点+判断单向连通)

题意:判断一个有向图中的任意两点u.v,是否可以由其中一个点到达另一个点. 分析:这个问题转化以后就是:将该图强连通缩点后再判断其是否是单向连通的.缩点用Tarjan处理强连通分量. 有一个定理是这样的:一个有向图是单项连通的当且仅当其拓扑排序唯一.那么将这个子问题再转化为其缩点之后的图拓扑排序是否唯一. 如果一个有向图拓扑排序唯一,那么在根据入度求拓扑排序的过程中,不会有超过一个点在同一时刻同时为0. #include<stack> #include<stdio.h> #incl

poj 2762 Going from u to v or from v to u?【强连通分量缩点+拓扑排序】

Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 15812 Accepted: 4194 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors

POJ 2762 Going from u to v or from v to u? Tarjan算法学习例题

Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17104 Accepted: 4594 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time,

POJ 2762Going from u to v or from v to u?(强联通 + 缩点 + 拓扑排序)

[题意]: 有N个房间,M条有向边,问能否毫无顾虑的随机选两个点x, y,使从①x到达y,或者,②从y到达x,一定至少有一条成立.注意是或者,不是且. [思路]: 先考虑,x->y或者y->x是什么意思,如果是且的话就简单了,就直接判断整个图是不是强联通图即可,但是这道题是或,那么可以随手画出一个DAG 比如1->3, 2->3 这样很明显是不行的,1,2没有联通,那么如果是这样1->2->3 就可以了,或者是1->2->3->1,这样也是可以的. 很

Oracle基本数据字典:v$database、v$instance、v$version、dba_objects

v$database: 视图结构: SQL> desc v$database; Name Null? Type ----------------------------------------- -------- ------------------------------------------------------------------------- DBID

POJ2762 Going from u to v or from v to u(单连通缩点)

判断图是否单连通,先用强连通分图处理,再拓扑排序,需注意: 符合要求的不一定是链拓扑排序列结果唯一,即在队列中的元素始终只有一个 #include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #in

Going from u to v or from v to u?_POJ2762强连通+并查集缩点+拓扑排序

Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each tim

临时文件相关的v$tempfile v$sort_usage与V$tempseg_usage

SQL> select username,user,segtype,segfile#,segblk#,extents,segrfno# from v$sort_usage; SEGFILE#代表的是绝对文件号(AFN),对应x$kcctf表中的TFAFN(temp file absolute file number). SQL> select indx,tfnum,tfafn,tfcsz from x$kcctf; TFCSZ 是创建临时文件的块数查看临时文件的绝对文件号 select tf

[强连通分量] POJ 2762 Going from u to v or from v to u?

Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17089 Accepted: 4590 Description In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors

POJ2762 Going from u to v or from v to u?（判定单连通图：强连通分量+缩点+拓扑排序）

这道题要判断一张有向图是否是单连通图,即图中是否任意两点u和v都存在u到v或v到u的路径. 方法是,找出图中所有强连通分量,强连通分量上的点肯定也是满足单连通性的,然后对强连通分量进行缩点,缩点后就变成DAG. 现在问题就变成,如何判断DAG是否是单连通图——用拓扑排序——如果拓扑排序过程中出现1个以上入度为0的点那就不是单连通图,因为有2个入度0的点,那这两个点肯定都无法到达对方. 另外,注意题目没说给的图是连通的!.. #include<cstdio> #include<cstrin

POJ 2762 Going from u to v or from v to u? (强连通分量缩点+拓扑排序)

题目链接:http://poj.org/problem?id=2762 题意是有t组样例,n个点m条有向边,取任意两个点u和v,问u能不能到v 或者v能不能到u,要是可以就输出Yes,否则输出No.注意一点,条件是或者!所以不是判断双连通图的问题. 我一开始没看到'or'这个条件,所以直接tarjan判断是否只有一个强连通分量,果断WA. 所以需要给原图缩点,用tarjan把图变成一个有向无环图,要是只有一个scc,那就直接输出Yes.那接下来讨论多个scc,要是新图中有两个及以上的点的入度为

[poj2762] Going from u to v or from v to u?(Kosaraju缩点+拓排)

转载请注明出处: http://www.cnblogs.com/fraud/ ——by fraud Going from u to v or from v to u? Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 14778 Accepted: 3911 Description In order to make their sons brave, Jiajia and Wind take t

关于oracle动态视图v$datafile和v$datafile_header(转)

v$datafile是从oracle的控制文件中获得的数据文件的信息v$datafile_header是从数据文件的头部在正常运行下,两者的检查点SCN值是一致的,但当datafile出现损坏时可以用v$datafile_header确定应用重做的起始SCN,而v$datafile可确定应用重做的结束SCN值sql> select a.name,a.checkpoint_change# “start_SCN”, > b.checkpoint_change# “last_SCN” &

v-if和v-else指令求苹果总价

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