1063. 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%


代码：

#include <cstdio>
#include <iostream>
#include <set>
using namespace std;
int main()
{
    int n,m,k,d,a,b;
    scanf("%d",&n);
    set<int> q[50];
    for(int i = 0;i < n;i ++)
    {
        scanf("%d",&m);
        for(int j = 0;j < m;j ++)
        {
            scanf("%d",&d);
            q[i].insert(d);
        }
    }
    scanf("%d",&k);
    for(int i = 0;i < k;i ++)
    {
        scanf("%d%d",&a,&b);
        int c = 0;
        for(set<int>::iterator j = q[a - 1].begin();j != q[a - 1].end();j ++)
        {
            if(q[b - 1].find(*j) != q[b - 1].end())c ++;
        }
        printf("%.1f%%
",c * 100.0 / (q[a - 1].size() + q[b - 1].size() - c));
    }
}