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…

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…

这题比较简单,注意路径压缩即可. AC代码 //#define LOCAL #include <stdio.h> #include <algorithm> using namespace std; const int maxn = 20000+5; int par[maxn], dis[maxn]; void init(int n) { for(int i = 0; i <= n; i++) { par[i] = i; dis[i] = 0; } } int findRoot…

·一些很可爱的询问和修改,放松地去用并查集解决. ·英文题,述大意: 输入n(5<=n<=20000)表示树有n个节点,并且会EOF结束地读入不超过 20000个操作,一共有两种:    ①I v u:表示将v的父亲节点设置为u(在这之前v没有爸爸),边权设置为abs(v-u)%1000.    ②E u:表示询问u到当前它所在树的根节点的距离. ·分析:      为了记录当前一系列加边操作后所有的点的位置情况(因为你随时可能回答询问啊),根据这道题的点关系特点[只在乎点和其根节点的信息],…

题目传送门 题意:训练指南P192 分析:主要就是一个在路径压缩的过程中,更新点i到根的距离 #include <bits/stdc++.h> using namespace std; const int N = 2e4 + 5; struct DSU { int rt[N], d[N]; void init(void) { memset (rt, -1, sizeof (rt)); memset (d, 0, sizeof (d)); } int Find(int x) { if (rt[x…

这题感觉和 POJ 1988 Cube Stacking 很像,在路径压缩的同时递归出来的时候跟新distant数组 我发现我一直WA的原因是,命令结束是以字母o结束的,而不是数字0!! //#define LOCAL #include <algorithm> #include <cstdio> using namespace std; + ; int parent[maxn], distant[maxn]; int GetParent(int a) { if(parent[a]…

UVAlive 3027 Corporative Network 题目:   Corporative Network Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 3450   Accepted: 1259 Description A very big corporation is developing its corporative network. In the beginning each of the N ent…

3027 - Corporative Network 思路:并查集: cost记录当前点到根节点的距离,每次合并时路径压缩将cost更新. 1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<string.h> 5 #include<math.h> 6 #include<stack> 7 #include<set> 8 #inc…

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…

                     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,…

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…

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…

题意:一个企业要去收购一些公司把,使的每个企业之间互联,刚开始每个公司互相独立 给出n个公司,两种操作 E I:询问I到I它连接点最后一个公司的距离 I I J:将I公司指向J公司,也就是J公司是I公司的上级,距离为abs(I-J)%1000(貌似G++不支持abs,PE了两发) 思路:转化一下题意就行了,首先刚开始的时候每个公司都是独立的,I操作就是并查集中合并操作,将I这课树并到J这个树上, E操作要求的东西就是 I到I的根节点的距离,先看一个没有路径压缩直接暴力的方法把.(本以为不会过的,…

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 its own computing and telecommunication center. Soon, for amelioration of the se…

Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 18066   Accepted: 7618 Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computer…

Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers…

An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers are repaired…

---恢复内容开始--- 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 ente…

设这个序列的前缀和为Si(0 <= i <= n),S0 = 0 每一个符号对应两个前缀和的大小关系,然后根据这个关系拓扑排序一下. 还要注意一下前缀和相等的情况,所以用一个并查集来查询. #include <cstdio> #include <cstring> ; int n; int G[maxn][maxn]; ]; int sum[maxn], a[maxn]; int topo[maxn], c[maxn], t; void dfs(int u) { c[u]…

链接:http://poj.org/problem?id=2236 题意: 有一个计算机网络,n台计算机全部坏了,给你两种操作: 1.O x 修复第x台计算机 2.S x,y 判断两台计算机是否联通 联通的条件: 两台都修复好了,求距离小于d 思路: 数据很小,我们可以没修复一台就直接枚举已经修复的计算机找到距离小于d的,放到一个并查集里,查询的话直接查询是否再同一个并查集里就好了 实现代码: //while(scanf("\n%c", &ope) != EOF) #inclu…

An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers are repaired…

水题 #include<cstdio> #include<cstring> #include<queue> #include<set> #include<cstdlib> #include<algorithm> #include<vector> #include<cmath> using namespace std; typedef long long LL; typedef pair<int,int&g…

转自:http://blog.csdn.net/shahdza/article/details/7779230 [HDU]1213 How Many Tables 基础并查集★1272 小希的迷宫 基础并查集★1325&&poj1308 Is It A Tree? 基础并查集★1856 More is better 基础并查集★1102 Constructing Roads 基础最小生成树★1232 畅通工程 基础并查集★2120 Ice_cream's world I 基础并查集★212…

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…

<题目链接> 题目大意: n个节点,若干次询问,I x y表示从x连一条边到y,权值为|x-y|%1000:E x表示询问x到x所指向的终点的距离.   解题分析: 与普通的带权并查集类似,但是要注意的是,在进行查询操作时,也要调用一下find函数,因为以前的uion操作可能将某些值改变了,所以find()操作是将这些节点全部更新一遍,得到它们的真实值. #include <cstdio> #include <cmath> +; int father[maxn],dis…

P.S.我不想看英文原题的,但是看网上题解的题意看得我 炒鸡辛苦&一脸懵 +_+,打这模版题的代码也纠结至极了......不得已只能自己翻译了QwQ . 题意:有一个公司有N个企业,分成几个网络,分别从各个网络中选一个机器设置为中心机.下面有2种操作:1.查询当前时间机器x到其所在网络的中心机的距离:2.设置中心机x与机器y相连,距离为abs(x-y)%1000,x所在的网络的中心机变为y所在网络的中心机. 解法:带权并查集.可以把中心机转换为一个集合(树)的根节点,求距离就是求点到根节点的距离…

Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 25022   Accepted: 10399 Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap compute…

传送门  Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 24513   Accepted: 10227 Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap co…

Wireless Network Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 16065   Accepted: 6778 Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computer…

Wireless Network 题目链接: http://acm.hust.edu.cn/vjudge/contest/123393#problem/A Description An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aft…