【算法笔记】A1063 Set Similarity

1063 Set Similarity （25 分)

 

Given two sets of integers, the similarity of the sets is defined to be /, 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 (≤) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤) and followed by M integers in the range [0]. After the input of sets, a positive integer K (≤) 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%

题意：

　　比较两个set<int>集合的相似度。

code：

 #include<bits/stdc++.h>
 using namespace std;
 vector<set<int> > sets;
 int main(){
     int n,m,k,a1,a2;
     scanf("%d",&n);
     sets.resize(n);
     for (int i = ; i < n; i++)
     {
         scanf("%d",&m);
         set<int> temp;
         for (int j=;j<m;j++)
         {
             int num;
             scanf("%d",&num);
             temp.insert(num);
         }
         sets[i]=temp;
     }
     scanf("%d",&k);
     for (int i=;i<k;i++)
     {
         scanf("%d %d",&a1,&a2);
         int nc=,nt = sets[a2-].size();
         for (auto it=sets[a1-].begin();it!=sets[a1-].end();it++)
         {
             if (sets[a2-].find(*it)==sets[a2-].end())
                 nt++;
             else
                 nc++;
         }
         printf("%.1lf%%\n",(double)nc/nt*);
     }
     return ;
 }