Corporative Network
Time Limit: 3000MS   Memory Limit: 30000K
Total Submissions: 3664   Accepted: 1326

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 services, the corporation started to collect some enterprises in clusters, each of them served by a single computing and telecommunication center as follow. The corporation chose one of the existing centers I (serving the cluster A) and one of the enterprises J in some other cluster B (not necessarily the center) and link them with telecommunication line. The length of the line between the enterprises I and J is |I – J|(mod 1000).In such a way the two old clusters are joined in a new cluster, served by the center of the old cluster B. Unfortunately after each join the sum of the lengths of the lines linking an enterprise to its serving center could be changed and the end users would like to know what is the new length. Write a program to keep trace of the changes in the organization of the network that is able in each moment to answer the questions of the users.

Input

Your program has to be ready to solve more than one test case. The first line of the input will contains only the number T of the test cases. Each test will start with the number N of enterprises (5<=N<=20000). Then some number of lines (no more than 200000) will follow with one of the commands:
E I – asking the length of the path from the enterprise I to its serving center in the moment;

I I J – informing that the serving center I is linked to the enterprise J.

The test case finishes with a line containing the word O. The I commands are less than N.

Output

The
output should contain as many lines as the number of E commands in all
test cases with a single number each – the asked sum of length of lines
connecting the corresponding enterprise with its serving center.

Sample Input

1
4
E 3
I 3 1
E 3
I 1 2
E 3
I 2 4
E 3
O

Sample Output

0
2
3
5
题意:有n个点,一开始每个点以自己为信号终点,当输入I x y,就连接x,y(len[x,y] =|x-y|%1000),并且X的信号终点指向Y,E x是输出x距离信号终点的距离,输O结束当前输入(题目好像有点
问题,N说要大于等于5,测试用例给的是4) 这题和我上面那篇博客一样的都是利用递归更新权值。值得注意的是这题询问的时候不用去寻找两个点的根结点,因为他们本来就是相邻结点了(父子关系)。
import java.util.Scanner;

public class Main {
final static int MAXSIZE = 20005;
static int[] father;
static int[] len; public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int tcase = sc.nextInt();
while (tcase-- > 0) {
int n = sc.nextInt();
father = new int[n+2];
len = new int[n+2];
for (int i = 1; i <= n; i++) {
father[i] = i;
len[i] = 0;
}
while (true) {
String str = sc.next();
if (str.equals("E")) {
int a = sc.nextInt();
find(a);
System.out.println(len[a]);
} else if(str.equals("I")){
int a = sc.nextInt();
int b = sc.nextInt();
father[a] = b;
len[a]=Math.abs(a-b)%1000;
} else break;
}
}
} private static int find(int a) {
if (a != father[a]) {
int temp = find(father[a]);
len[a] += len[father[a]];
father[a] = temp;
}
return father[a];
}
}

poj 1962(并查集+带权更新)的更多相关文章

  1. 浅谈并查集&种类并查集&带权并查集

    并查集&种类并查集&带权并查集 前言: 因为是学习记录,所以知识讲解+例题推荐+练习题解都是放在一起的qvq 目录 并查集基础知识 并查集基础题目 种类并查集知识 种类并查集题目 并查 ...

  2. POJ 1182 食物链 [并查集 带权并查集 开拓思路]

    传送门 P - 食物链 Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit  ...

  3. 种类并查集——带权并查集——POJ1182;HDU3038

    POJ1182 HDU3038 这两个题比较像(一类题目),属于带权(种类)并查集 poj1182描绘得三种动物种类的关系,按照他一开始给你的关系,优化你的种类关系网络,最后看看再优化的过程中有几处矛 ...

  4. Poj1182 食物链(并查集/带权并查集)

    题面 Poj 题解 这里采用并查集的补集. \(x\)表示同类集合,\(x+n\)表示敌人集合,\(x+n\times2\)表示敌人的敌人集合. 如果当前给出的是一对同类关系,就判断\(x\)是否吃\ ...

  5. HDU 3038 How Many Answers Are Wrong 并查集带权路径压缩

    思路跟 LA 6187 完全一样. 我是乍一看没反应过来这是个并查集,知道之后就好做了. d[i]代表节点 i 到根节点的距离,即每次的sum. #include <cstdio> #in ...

  6. 【并查集&&带权并查集】BZOJ3296&&POJ1182

    bzoj1529[POI2005]ska Piggy banks [题目大意] n头奶牛m种语言,每种奶牛分别掌握一些语言.问至少再让奶牛多学多少种语言,才能使得它们能够直接或间接交流? [思路] ( ...

  7. POJ 1988 Cube Stacking( 带权并查集 )*

    POJ 1988 Cube Stacking( 带权并查集 ) 非常棒的一道题!借鉴"找回失去的"博客 链接:传送门 题意: P次查询,每次查询有两种: M x y 将包含x的集合 ...

  8. POJ 2195 Going Home (带权二分图匹配)

    POJ 2195 Going Home (带权二分图匹配) Description On a grid map there are n little men and n houses. In each ...

  9. poj 1984 并查集

    题目意思是一个图中,只有上下左右四个方向的边.给出这样的一些边, 求任意指定的2个节点之间的距离. 就是看不懂,怎么破 /* POJ 1984 并查集 */ #include <stdio.h& ...

随机推荐

  1. SRM13 T3 花六游鸟小(结论题)

    哇这题是真的喵,HR智商太高辣 这题的难点就是看了题解之后怎么证明题解里的结论... 结论①:深度大于logm的点肯定能达到最大值 证明:显然一个西瓜的属性里0数量一半1数量一半我们取到的1数量最少, ...

  2. 34张史上最全IT架构师技术知识图谱 最新下载

    本文是笔者多年来积累和收集的知识技能图谱,小编极力推荐分享给身边的技术人儿,希望这份技术知识图谱能够帮助到每一位奋斗在技术路上的小伙伴. 下面是笔者多年来积累和收集的知识技能图谱,有的是笔者原创总结的 ...

  3. HDU 5640

    King's Cake Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total ...

  4. Codeforces Round #202 (Div. 2) B,C,D,E

    贪心 B. Color the Fence time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  5. c# 定时执行任务

    在Global.asax文件中加上 void Application_Start(object sender, EventArgs e) { // Code that runs on applicat ...

  6. MongoDB入门(6)- 我们自己封装的MongoDB-C#版本

    Wisdombud.Mongo 包含内容 MongoDB.Bson.dll MongoDB.Bson.xml MongoDB.Driver.dll MongoDB.Driver.xml Wisdomb ...

  7. Mybatis 参考

    1:Mybatis最入门---ResultMaps基本用法 2:Mybatis最入门---ResultMaps高级用法(上) 3:Mybatis最入门---ResultMaps高级用法(下) 4:My ...

  8. 使用pipenv管理python项目

    入门 首先使用pip安装Pipenv及其依赖项, pip install pipenv 1 然后将目录更改为包含你的Python项目的文件夹,并启动Pipenv, cd my_project pipe ...

  9. Spring Security 集成CAS实现单点登录

    参考:http://elim.iteye.com/blog/2270446 众所周知,Cas是对单点登录的一种实现.本文假设读者已经了解了Cas的原理及其使用,这些内容在本文将不会讨论.Cas有Ser ...

  10. 【ALB学习笔记】基于多线程方式的串行通信接口数据接收案例

    基于多线程方式的串行通信接口数据接收案例 广东职业技术技术学院  欧浩源 1.案例背景 在本博客的<[CC2530入门教程-06]CC2530的ADC工作原理与应用>中实现了电压数据采集的 ...