1063. Set Similarity

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.

最后一个测试点超时，布吉岛

#include <stdio.h>
#include <stdlib.h>
#include <set>
using namespace std;
#define N 51

int main ()
{
    int n,i,j,m;
    scanf("%d",&n);
    set<long> num[n];
    long temp;
    for( i=0;i<n;i++)
    {
         scanf("%d",&m);
         for( j=0;j<m;j++)
         {
              scanf("%ld",&temp);
              num[i].insert(temp);
              }
         }
    int k,a,b;
    scanf("%d",&k);
    int len1,len2,len3;
    set<long> sum;
    for( i=0;i<k;i++)
    {
         sum.clear();
         scanf("%d %d",&a,&b);
         len1=num[a-1].size();
         len2=num[b-1].size();
         for( set<long>::iterator it=num[a-1].begin();it!=num[a-1].end();it++)
              sum.insert(*it);
         for( set<long>::iterator it=num[b-1].begin();it!=num[b-1].end();it++)
              sum.insert(*it);
         len3=sum.size();
         float rate=(len1+len2-len3)*1.0f/len3*100.0f;
         printf("%.1f",rate);
         printf("%%\n");
         }

    system("pause");
    return 0;
    }