A1063. Set Similarity (25)

Given two sets of integers, the similarity of the sets is defined to be N_c/N_t*100%, where N_c is the number of distinct common numbers shared by the two sets, and N_t 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%

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <set> 
using namespace std;
const int N=51;
set<int> st[N];

void compare(int x,int y)
{
    int total=st[x].size(),same=0;
    for(set<int>::iterator it=st[y].begin();it!=st[y].end();it++)
    {
        if(st[x].find(*it)!=st[x].end())
        {
          same++; 
        } else total++;
    } 
    printf("%.1f%%
",same*100.0/total);
}
 
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int m;
        scanf("%d",&m);
        for(int j=0;j<m;j++)
        {
            int temp;
            scanf("%d",&temp);
            st[i].insert(temp);
        }
    }
    int k;
    scanf("%d",&k);
    for(int i=0;i<k;i++)
    {
        int a,b;
        scanf("%d %d",&a,&b);
        compare(a,b);
    }
    
    return 0;
}