题目原文描述:

Given a social network containing. n members and a log file containing m timestamps at which times pairs of members formed friendships, design an algorithm to determine the earliest time at which all members are connected (i.e., every member is a friend of a friend of a friend ... of a friend). Assume that the log file is sorted by timestamp and that friendship is an equivalence relation. The running time of your algorithm should be mlogn or better and use extra space proportional to n.

分析:

题目的意思是有一个包含n个成员的社交网络,日志文件log按照时间戳顺序存储了两个成员之间成为朋友的时间,共有m条记录。让我们设计一个算法来根据这个log文件来计算m个成员全部通过朋友关系连通的时间。

这是个典型的并查集。思路是读取日志文件,遍历文件记录,逐条记录union。采用加权quick-union算法,就可以满足mlogn的复杂度要求。作业提交100

 import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.util.Scanner; import edu.princeton.cs.algs4.StdOut;
import edu.princeton.cs.algs4.WeightedQuickUnionUF; public class SocialNetworkConnUF {
private FileInputStream ins;
private WeightedQuickUnionUF uf;
public SocialNetworkConnUF(int num, FileInputStream ins){
this.ins = ins;
uf = new WeightedQuickUnionUF(num);
} @SuppressWarnings("resource")
public String getEarliestConTime(){
Scanner scanner = new Scanner(ins,"utf-8");
String earliestConTime = null;
while(scanner.hasNextLine()){
String line = scanner.nextLine();
if(line != null && !line.trim().equals("")){
String[] lineArray = line.split(" ");
if(lineArray.length == 3){
String timestamp = lineArray[0];
int p = Integer.parseInt(lineArray[1]);
int q = Integer.parseInt(lineArray[2]);
if(uf.connected(p, q)) continue;
uf.union(p,q);
if(uf.count() == 1) {
earliestConTime = timestamp;
break;
}
}
} }
return earliestConTime;
}
public static void main(String[] args){
FileInputStream ins;
try {
ins = new FileInputStream("socialNetworkLog.txt");
SocialNetworkConnUF socialNet = new SocialNetworkConnUF(10, ins);
String earliestConnTime = socialNet.getEarliestConTime();
StdOut.println(" the earliest connected time is :" + earliestConnTime);
} catch (FileNotFoundException e) {
e.printStackTrace();
} }
/*
* socialNetworkLog.txt
* 20170714001 0 1
* 20170714002 4 5
* 20170714003 8 9
* 20170714004 2 4
* 20170714005 5 6
* 20170714006 7 8
* 20170714007 2 5
* 20170714008 6 7
* 20170714009 1 2
* 20170714010 0 3
* 20170714011 1 9
* 20170714012 3 7
*
*/
}

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