面对浩瀚的信息海洋，找到想要的资源有时真的是不容易。在大量文字中搜索高频词汇是信息搜索和数据压缩的共通课题。
这次智慧擂台请大家在一个比较庞大的英文文本中找出M个数量最多的短语（由N个单词组成）。统一处理相同的文本文件，该文本只包含英文单词、空格和回行符，比较谁的程序效率最高。
　　程序输入：M，N，文本文件路径（M不超过20，N不超过8）
程序输出：高频短语及其数量清单

擂台规则：提交符合以上要求的可执行程序，语言招式不限，点到为止；
我们将在统一的环境下对每位选手的作品进行公平的测试，
比较出综合用时最少的程序。
源程序

import java.io.*;
import java.util.*;

class tt
{
    public String phrase;
    public int count;
}

public class searchphrase
{

    private static LinkedHashMap phrase = new LinkedHashMap();
    static tt[] max_phrase;

    private static Vector SeparateString(String s)
    {
        Vector v = new Vector();
        String temp = "";
        for (int i = 0; i < s.length(); i++)
        {
            if (s.charAt(i) != ' ')
            {
                temp += s.charAt(i);
            }
            else
            {
                if (temp != "")
                    v.add(temp);
                temp = "";
            }
        }
        if (temp != "")
            v.add(temp);
        return v;
    }

    private static void swap(int pos, int count, String phrase)
    {
        int i;
        if (max_phrase[pos - 1].count < count)
        {
            for (i = pos - 1; i > 0; i--)
            {
                if (max_phrase[i - 1].count > max_phrase[i].count)
                    break;
            }
            max_phrase[pos].count = max_phrase[i].count;
            max_phrase[pos].phrase = max_phrase[i].phrase;
            max_phrase[i].count = count;
            max_phrase[i].phrase = phrase;
        }

    }

    private static void adjust_max(int count, String phrase)
    {
        int i, j;
        if (count <= max_phrase[max_phrase.length - 1].count)
            return;
        for (i = max_phrase.length - 1; i >= 0; i--)
        {
            if (max_phrase[i].phrase.equals(phrase))
            {
                max_phrase[i].count = count;
                if (i > 0)
                {
                    swap(i, count, phrase);
                }
                return;
            }
        }
        max_phrase[max_phrase.length - 1].count = count;
        max_phrase[max_phrase.length - 1].phrase = phrase;
        if (i > 0)
        {
            swap(max_phrase.length - 1, count, phrase);
        }
    }

    private static void js(Vector v, int n)
    {
        String s;
        for (int i = 0; i < v.size() - n + 1; i++)
        {
            s = "";
            for (int j = i; j < i + n; j++)
            {
                s += v.get(j) + " ";
            }
            int count = 1;
            if (phrase.containsKey(s.hashCode()))
            {
                count = Integer.parseInt(phrase.get(s.hashCode()).toString());
                count++;
            }
            phrase.put(s.hashCode(), count);
            adjust_max(count, s);
        }
    }

    public static void main(String[] args)
    {
        try
        {
            long t;
            int m, n;
            String path;
            m = Integer.parseInt(args[0]);
            n = Integer.parseInt(args[1]);
            path = args[2];
            max_phrase = new tt[m];
            for (int i = 0; i < m; i++)
            {
                max_phrase[i] = new tt();
                max_phrase[i].count = 0;
                max_phrase[i].phrase = "";
            }
            t = (new java.util.Date()).getTime();
            java.io.FileReader fr = new java.io.FileReader(path);
            java.io.BufferedReader br = new BufferedReader(fr);
            String s;

            Vector v = null;

            while ((s = br.readLine()) != null)
            {
                v = SeparateString(s);
                js(v, n);
            }
            for (int i = 0; i < m; i++)
            {
                System.out.println(max_phrase[i].phrase);
                System.out.println(max_phrase[i].count);
                System.out.println();
            }
            t = (new java.util.Date()).getTime() - t;
            System.out.print(t);
            System.out.println(" ms");
        }
        catch (Exception e)
        {
            System.out.println(e.getMessage());
        }

    }

}
测试结果1：m = 20 n = 8 
under games played won drawn lost goals for
71
tabulated under games played won drawn lost goals
70
games played won drawn lost goals for against
70
May Xinhua Following are the results from the
69
played won drawn lost goals for against and
59
won drawn lost goals for against and points
59
Jan Xinhua Following are the results from the
48
Chinas economic efficiency indicators of the sector of
39
The industrial statistics include all stateowned enterprises and
39
industrial statistics include all stateowned enterprises and the
39
statistics include all stateowned enterprises and the nonstateowned
39
include all stateowned enterprises and the nonstateowned ones
39
all stateowned enterprises and the nonstateowned ones with
39
stateowned enterprises and the nonstateowned ones with annual
39
enterprises and the nonstateowned ones with annual sales
39
and the nonstateowned ones with annual sales income
39
Xinhua Chinas economic efficiency indicators of the sector
39
the nonstateowned ones with annual sales income over
39
nonstateowned ones with annual sales income over million
39
up percent over the same period last year
35
13594 ms

测试结果2  m = 10 n = 5
Xinhua Following are the results
295
May Xinhua Following are the
209
Following are the results from
183
are the results from the
176
April Xinhua Following are the
141
Jan Xinhua Following are the
122
billion yuan billion US dollars
120
won drawn lost goals for
88
played won drawn lost goals
88
Dec Xinhua Following are the
87
12437 ms

以上源程序是采用的是最简单的方法，谁有更好，效率更高的方法，请跟贴!!