求集合內兩兩字符串的編輯距離，先使用暴力方法，之后再介紹用kd樹的方法

主線程進行任務分發，將字符串依次放入線程池隊列中，線程池內的線程就計算這個字符串和所有字符串的編輯距離，然后將計算結果寫入blockqueue中，再起一個線程將blockqueue中的結果寫入磁盤中.這里有一個技巧，當處理完所有字符串，寫進程可能在將隊列中的結果寫入到磁盤也可能阻塞在take函數處。主線程就在隊列末尾放入一個結束標記，讀線程接收到這個標記之后就拋出異常結束了。

import java.io.BufferedReader; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.List; import java.util.PriorityQueue; import java.util.Scanner; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import java.util.concurrent.LinkedBlockingQueue; import java.util.concurrent.TimeUnit;public class EditDistance{ExecutorService service ;List<String> ls;LinkedBlockingQueue<String> outque;class WordDist implements Comparable{public String word;public int dist;public WordDist(String wd, int d) {word = wd;dist = d;}public int compareTo(Object o) {return dist - ((WordDist)o).dist ;}public String toString () {return word + " " + dist;}}class Process implements Runnable {String name;public Process(String s) {name = s;}public void run() {PriorityQueue<WordDist> que = new PriorityQueue<WordDist>();for (int j = 0; j < ls.size(); ++j) {if (!name.equals(ls.get(j))) {que.add(new WordDist(ls.get(j), calcEditDist(name, ls.get(j))));}}StringBuilder sb = new StringBuilder();sb.append(name+"\t");int num = 20;int count = 0;while (que.peek() != null && count < num) {sb.append(que.poll() + " ");++count;}try {outque.put(sb.toString());} catch (InterruptedException e) {System.out.println(Thread.currentThread().getName() + " " + name);e.printStackTrace();}}}class Writer implements Runnable {PrintWriter pw;public Writer(String fn) {try {pw = new PrintWriter(fn);} catch (IOException e) {e.printStackTrace();return;}}@Overridepublic void run() {try {while (true) {String rst = outque.take();if (rst.equals("!!!POISON")) {throw new InterruptedException("meet a poison. over");}pw.println(rst);}} catch (InterruptedException e) {pw.close();System.out.println(e);System.out.println("Writer is over");}}}public int calcEditDist(String s1, String s2) {int [][]mat = new int[s1.length()+1][s2.length()+1];//mat[0][0] = 0;for (int i = 1; i <= s1.length(); ++i) mat[i][0] = i;for (int i = 1; i <= s2.length(); ++i) mat[0][i] = i;for (int i = 1; i <= s1.length(); ++i) {for (int j = 1; j <= s2.length(); ++j) {int repCost = 1;if (s1.charAt(i-1) == s2.charAt(j - 1)) {repCost = 0;}mat[i][j] = min(mat[i-1][j-1]+repCost, mat[i-1][j] + 1, mat[i][j-1] + 1);}}return mat[s1.length()][s2.length()];}private int min(int a, int b, int c) {int m = a;if (m > b) m = b;if (m > c) m = c;return m;}public void genCluster(String in, String out) {try {BufferedReader br = new BufferedReader (new InputStreamReader(new FileInputStream (in) ,"utf-8"));//PrintWriter pw = new PrintWriter(out, "utf-8");service = Executors.newFixedThreadPool(7);String line;ls = new ArrayList<String>();outque = new LinkedBlockingQueue<String>();//while ((line = br.readLine()) != null) {Scanner scan = new Scanner(line);scan.useDelimiter(" \\|\\|\\| ");ls.add(scan.next());}br.close();Thread writer = new Thread(new Writer(out));writer.start();for (int i = 0; i < ls.size(); ++i) {service.submit(new Process(ls.get(i)));}service.shutdown();//提交完任務之后就調用shutdown,不能再提交新任務try {service.awaitTermination(24, TimeUnit.HOURS);//再接著阻塞等待所有任務的結束} catch (InterruptedException e) {e.printStackTrace();System.out.println("main await error");}try {outque.put("!!!POISON");} catch (InterruptedException e) {e.printStackTrace();System.out.println("put poison error");}} catch (IOException e) {e.printStackTrace();}}public static void main(String[] args) {// TODO Auto-generated method stubif (args.length != 2) {System.out.println("in out");return;}EditDistance dist = new EditDistance();dist.genCluster(args[0], args[1]);}}
service.awaitTermination(24, TimeUnit.HOURS);

這個函數出問題了，java多線程和c有個不同之處在于，只要有一個線程再跑著，只要不是后臺線程，即使main線程執行結束了，就是用jstack看不到main線程了，其他線程也會繼續執行

恩，上面的程序跑了不止24小時，main線程執行結束了，也用poison把寫線程結束了，只剩下生產者線程了.....

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