PAT甲级A1063 Set Similarity （25 分）

最新推荐文章于 2020-05-27 16:30:21 发布

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最新推荐文章于 2020-05-27 16:30:21 发布

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本文介绍了一种计算两个集合相似度的方法，定义为Nc/Nt*100%，其中Nc为两个集合中相同且不同的数字数量，Nt为两个集合中所有不同数字的总数。文章提供了使用C++和unordered_map实现的示例代码。

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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%

题意：给出若干个数字集合，然后定义集合相似度为Nc/Nt*100%，其中 Nc为两个集合中相同数字的总个数，要求Nc中每个数字均不同，Nt为两个集合中不同数字的总个数。

思路：使用unordered_map存储每个集合，然后再计算相同个数。

参考代码：

#include<cstdio>
#include<unordered_map>
using namespace std;
int main()
{
	int n,num,t,q,id1,id2;
	scanf("%d",&n);
	unordered_map<int,bool> mp[52];
	for(int i=1;i<=n;i++){
		scanf("%d",&num);
		for(int j=0;j<num;j++){
			scanf("%d",&t);
			mp[i][t]=true;
		}
	}
	scanf("%d",&q);
	for(int i=0;i<q;i++){
		scanf("%d%d",&id1,&id2);
		int Nc=0,Nt;
		for(auto it=mp[id1].begin();it!=mp[id1].end();it++){
			if(mp[id2].find(it->first)!=mp[id2].end())
				Nc++;
		}
		Nt=mp[id1].size()+mp[id2].size()-Nc;
		printf("%.1f%%\n",(float)Nc/Nt*100);
	}
	return 0;
}