Set Similarity (25)

最新推荐文章于 2021-10-04 15:08:41 发布

原创最新推荐文章于 2021-10-04 15:08:41 发布 · 315 阅读

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本文介绍了一个计算两个整数集合相似度的问题，通过编程实现来计算两个集合共有元素占总不重复元素的比例，并提供了C++和Java两种语言的实现代码。

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题目描述

Given two sets of integers, the similarity of the sets is defined to be Nc/Nt*100%, where Nc is the number of distinct common numbers shared by the two sets, and Nt is the total number of distinct numbers in the two sets.  Your job is to calculate the similarity of any given pair of sets.

输入描述:

Each input file contains one test case.  Each case first gives a positive integer N (<=50) which is the total number of sets.  Then N lines follow, each gives a set with a positive M (<=104) and followed by M integers in the range [0, 109].  After the input of sets, a positive integer K (<=2000) is given, followed by K lines of queries.  Each query gives a pair of set numbers (the sets are numbered from 1 to N).  All the numbers in a line are separated by a space.

输出描述:

For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.

输入例子:

3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3

输出例子:

50.0%
33.3%

我的代码（C++）：

#include<iostream>
#include<set>
using namespace std;
int main()
{
    set<int> s[50];
    set<int>::iterator it;
    int x,y,z,n,m1,m2,i;
    scanf("%d",&x);
    for(i=0;i<x;i++)
    {
        scanf("%d",&y);
        while(y--)
        {
            scanf("%d",&z);
            s[i].insert(z);
        }
    }
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d",&m1,&m2);
        int cnt=0;
        for(it=s[m1-1].begin();it!=s[m1-1].end();it++)
        {
            if(s[m2-1].find(*it)!=s[m2-1].end()) cnt++;
        }
        double sum=s[m1-1].size()+s[m2-1].size()-cnt;
        printf("%.1f%%\n",cnt/sum*100);
    }
    return 0;
}

我的代码（Java）：

import java.util.HashSet;
import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner In=new Scanner(System.in);
		int x=In.nextInt();
		HashSet[] s=new HashSet[x];
		for(int i=0;i<x;i++)
		{
			int y=In.nextInt();
			s[i]=new HashSet<Integer>(y);
			for(int j=0;j<y;j++)
			{
				int z=In.nextInt();
				s[i].add(z);
			}
		}
		int n=In.nextInt();
		for(int i=0;i<n;i++)
		{
			int m1=In.nextInt();
			int m2=In.nextInt();
			int cnt=0;
			HashSet<Integer> s1=s[m1-1];
			HashSet<Integer> s2=s[m2-1];
			for(Integer integer:s2)
			{
				if(s1.contains(integer)) cnt++;
			}
			double sum=s1.size()+s2.size()-cnt;
			System.out.printf("%.1f%%\n",cnt/sum*100);
		}
	}
}