[字符串]Very easy

最新推荐文章于 2019-10-24 10:04:15 发布

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本文介绍了一个使用Java实现的ACM竞赛题目解决方案，该方案通过计算两个文档之间的相似度来判断它们是否足够相似。具体实现包括读取输入、过滤不重要的词汇，并采用余弦相似度的方法进行比较。

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Input

The test start with n which means that there are n cases.

The first line of each case gives two number p,q:the threshold and the size of excepted-word set.

The next line will give q words that you should ignore.

The next two lines are two document.

The document contains only words and all letters are lowercase.

The words are separated by blank space.

Output

If the two given documents are similar,print ’Yes’,or print ‘No’.

Sample Input

20.8 3i am ai am a boyi am a boy0.8 3i am ai am a boyi am a girl

Sample Output

YesNo

HINT

Source

第一次用Java做acm。

输入输出使用Scanner，配合String的split。非常方便

package A;
import java.util.*;
//import java.math.*;

public class Main {

	public static void main(String[] arvs)
	{
		Scanner a = new Scanner(System.in);
		int T = a.nextInt();
		while ((T--) > 0)
		{
			if (!a.hasNextDouble()) break;
			double p0 = a.nextDouble();
			int n = a.nextInt();
			HashSet<String> filter = new HashSet<String>();
			HashMap<String, Integer> hash1 = new HashMap<String,Integer>();
			HashMap<String, Integer> hash2 = new HashMap<String,Integer>();
			for (int i=0;i<n;i++)
			{
				String f = a.next(); 
				filter.add(f);
			}
			String line,words[];
			while ((line = a.nextLine()).equals(""));
			words = line.split(" ");
			for (String w:words)
			{
				if (filter.contains(w))
					continue;
				if (hash1.containsKey(w))
					hash1.put(w, hash1.get(w)+1);
				else
					hash1.put(w, 1);
			}
			
			double fenmu1 = 0;
			for (int frequence:hash1.values())
			{
				fenmu1 += frequence*frequence;
			}
			fenmu1 = Math.sqrt(fenmu1);
			
			while ((line = a.nextLine()).equals(""));
			words = line.split(" ");
			
			int common = 0;
			for (String w:words)
			{
				if (filter.contains(w))
					continue;
				if (hash1.containsKey(w))
				{
					common += hash1.get(w);
				}
				
				if (hash2.containsKey(w))
					hash2.put(w, hash2.get(w)+1);
				else
					hash2.put(w, 1);
			}
				
			double fenmu2 = 0;
			for (int frequence:hash2.values())
			{
				fenmu2 += frequence*frequence;
			}
			fenmu2 = Math.sqrt(fenmu2);
			
			double p = (double)common/(fenmu1*fenmu2);
			if (p < p0)
				System.out.println("No");
			else
				System.out.println("Yes");
		}
	}
}