1063 Set Similarity (25 分) java 题解

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.

Input Specification:

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.

Output Specification:

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

Sample Input:

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

Sample Output:

50.0%
33.3%

解题思路：

题目大意：输入某些集合，求指定两个集合中的并集：交集。

先将所有集合元素存储至二维数组中，求并集时，不仅要保证和指定集合有重合，还要保证和自身集合无重合。交集只需把两集合元素都放入同一set中即可。

注意局部变量的使用和输出的格式。

java代码：

import java.io.*;
import java.util.*;

public class Main {
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		int [][]arr = new int[n][];
		for(int i = 0; i < n;i++) {
			String[] split = br.readLine().split(" ");
			int m = Integer.parseInt(split[0]);
			arr[i] = new int[m];
			for(int j = 0; j < m;j++) {
				arr[i][j] = Integer.parseInt(split[j + 1]);
			}
		}
		int count = Integer.parseInt(br.readLine());
		StringBuilder builder = new StringBuilder();
		for(int i = 0; i < count;i++) {
			int ans = 0;
			String[] split = br.readLine().split(" ");
			int a = Integer.parseInt(split[0]) - 1;
			int b = Integer.parseInt(split[1]) - 1;
			Set set = new HashSet<>();
			Set set1 = new HashSet<>();
			for(int j = 0; j < arr[a].length;j++) {
				set.add(arr[a][j]);
			}
			for(int j = 0; j < arr[b].length;j++) {
				boolean key = set.contains(arr[b][j]);
				boolean key1 = set1.contains(arr[b][j]);
				if(key && !key1) {
					ans++;
				}
				set1.add(arr[b][j]);
			}
			set.addAll(set1);
			String str = String.format("%.3f",(ans * 1.0) / set.size());
			builder.append(String.format("%.1f%%n",Float.parseFloat(str) * 100));
//			System.out.printf("%.1f%%n",Float.parseFloat(str) * 100);
		}
		System.out.print(builder.toString().trim());
	}
}

PTA提交截图：