A very big corporation is developing its corporative network. In the beginning each of the N enterprises of thecorporation, numerated from 1 to N, organized its own computing and telecommunication center. Soon, foramelioration of the services, the co…
Corporative Network Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description   A very big corporation is developing its corporative network. In the beginning each of the N enterprises of th…
                     Corporative Network A very big corporation is developing its corporative network. In the beginning each of the N enterprisesof the corporation, numerated from 1 to N, organized its own computing and telecommunication center.Soon,…
题意: 有 n 个节点,初始时每个节点的父节点都不存在,你的任务是执行一次 I 操作 和 E 操作,含义如下: I  u  v   :  把节点 u  的父节点设为 v  ,距离为| u - v | 除以 1000 的余数. E u   : 询问u 到根节点的距离. 解题思路: 因为题目只查询节点到根节点的距离,所以每棵树处理根节点不能换之外.其他节点的位置可以随意改变,这恰好符合并查集的特点,但是附加了一点东西....在两点之间有了一个附加的权值(距离)..所以就是加权并查集了..题目给的是节…
1.无法操作根节点数据,提示的是This base cannot be created with PLA. 解决办法 1)添加一个base.ldif文件,里面的dc和配置文件里的保持一致即可 dn: dc=dc,dc=com o: ldap objectclass: dcObject objectclass: organization 2)然后执行命令导入数据库 ldapadd -f base.ldif -x -D cn=root,dc=dc,dc=com -W 然后会提示输入密码,添加成功会提…
A very big corporation is developing its corporative network. In the beginning each of the N enterprises of the corporation, numerated from 1 to N, organized its own computing and telecommunication center. Soon, for amelioration of the services, the…
3027 - Corporative Network A very big corporation is developing its corporative network. In the beginning each of the N enterprisesof the corporation, numerated from 1 to N, organized its own computing and telecommunication center.Soon, for ameliorat…
说明: 最开始是用的rehat7.2安装12.2.0.1,后面安装GI节点一执行root.sh脚本失败,排查原因,最开始以为是操作系统的问题,换成rehat7.6,同样的出现问题,经过一番折腾,后面通过MOS文献解决问题(如果直接18C,可以避免此BUG). 一.前期规划 1.软件环境 操作系统:RedHat Linux 7.x 64位 数 据 库:Oracle 12.2.0.1 RAC 2.主机规划 二.安装数据库前准备 1.清除 DNS 中内容 vi /etc/resolv.conf    …
Corporative Network Problem's Link Mean: 有n个结点,一开始所有结点都是相互独立的,有两种操作: I u v:把v设为u的父节点,edge(u,v)的距离为abs(u-v)%1000; E u:输出u到根节点的距离. analyse: 经典的并查集习题!Rujia挑选的题目真心不错. 做法很巧妙,但是要对并查集和路径压缩有深入的了解才能想到这种做法. 由于每次E操作都会加入一个新的结点,而且每次都是把一个结点指向另一个结点,所以说不会同时存在两个或两个以上…
Time Limit: 3000MS Memory Limit: 30000K Total Submissions: 3943 Accepted: 1414 Description A very big corporation is developing its corporative network. In the beginning each of the N enterprises of the corporation, numerated from 1 to N, organized i…