【PTA】 L2-005 Set similarity (25 points) 【set】

Given two sets of integers, their similarity is defined as: N​c /N​t ×100%. Where N​c
​​ is the number of unequal integers shared by the two sets, and N​t​​ is the number of unequal integers shared by the two sets. Your task is to calculate the similarity of any pair of given sets.

Input format: The
first line of input gives a positive integer N (≤50), which is the number of sets. Next N rows, each row corresponds to a collection. Each set is first given a positive integer M (≤10​4), which is the number of elements in the set; then followed by M integers in the interval [0,10​9​​].

The next line gives a positive integer K (≤2000), and then K lines, each line corresponds to the number of a pair of sets that need to calculate the similarity (the sets are numbered from 1 to N). The numbers are separated by spaces.

Output format:
For each pair of sets that need to be calculated, output their similarity in one line, which is a percentage number with 2 decimal places.

Input sample:

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:

50.00%
33.33%

---

Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m,k,x,a,b; 
set<int> g[60]; 
int main()
{
    
    	    
	cin>>n;
	for(int i=1;i<=n;i++)
	{
    
    
		cin>>m;
		for(int j=0;j<m;j++)
		{
    
    
			cin>>x;
			g[i].insert(x);
		}
	}
	
	cin>>k;
	for(int i=0;i<k;i++)
	{
    
    
		int cnt=0;
		cin>>a>>b;
		for(auto it=g[a].begin();it!=g[a].end();it++) 
			if(g[b].find(*it)!=g[b].end()) cnt++;  //Nc
		 
		double ans=cnt*100.0/(g[a].size()+g[b].size()-cnt);
		printf("%.2f%%\n",ans);
	}
	return 0;
}