1063 Set Similarity （25 分）

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 (≤10^​4​​ ) and followed by M integers in the range [0,10^​9​​ ]. 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%

题目大意：
本题定义集合相似度Nc/Nt*100%,其中Nc是两个集合中共有的不相等元素个数，Nt是两个集合中不相等元素的个数。你的任务是计算给定的集合的相似度。

#include <iostream>
#include <vector>
#include <set>
using namespace std;

int main(){
    int n;
    cin>>n;
    vector<set<int> > v(n);
    for(int i=0;i<n;i++){
        int m;
        cin>>m;
        while(m--){
            int val;
            cin>>val;
            v[i].insert(val);
        } 
    }
    int k;
    cin>>k;
    for(int i=0;i<k;i++){
        int a,b;
        cin>>a>>b;
        a--,b--;
        int cnt=0;
        for(auto it:v[a]){
            if(v[b].find(it)!=v[b].end()) cnt++;
        }
        printf("%.1lf%\n",cnt*100.0/(v[a].size()+v[b].size()-cnt));
    }
    return 0;
}