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%

 1 #include <cstdio>
 2 #include <set>
 3 #include <algorithm>
 4 #include <iterator>
 5
 6 using namespace std;
 7
 8 set<int> s[51];
 9
10 int main()
11 {
12     int setNum;
13     //cin >> setNum;
14     scanf("%d", &setNum);
15     for (int i = 1; i <= setNum; i++)
16     {
17         int keyNum;
18         scanf("%d", &keyNum);
19         //cin >> keyNum;
20         for (int j = 0; j < keyNum; j++)
21         {
22             int key;
23             scanf("%d", &key);
24             s[i].insert(key);
25         }
26     }
27     int queryNum;
28     scanf("%d", &queryNum);
29     //cin >> queryNum;
30     for (int i = 0; i < queryNum; i++)
31     {
32         int index1, index2;
33         scanf("%d%d", &index1, &index2);
34         int NC = 0;
35         set<int>::iterator it1, it2;
36         it1 = s[index1].begin();
37         it2 = s[index2].begin();
38         while (it1 != s[index1].end() && it2 != s[index2].end())
39         {
40             if (*it1 == *it2)
41             {
42                 it1++; it2++; NC++;
43             }
44             else if (*it1 < *it2)
45                 it1++;
46             else
47                 it2++;
48         }
49         int NP = s[index1].size() + s[index2].size() - NC;
50
51         printf("%.1f%%\n", NC*100.0 / NP);
52     }
53 }