A1063

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%

#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
set<int>s[55];
int main(){
    int n,k,x,m,a,b;
    cin>>n;
    for(int i=1;i<=n;i++){
        scanf("%d",&k);
        for(int j=0;j<k;j++){
            scanf("%d",&x);
            s[i].insert(x);
        } 
    }
    cin>>m;
    while(m--){
        scanf("%d%d",&a,&b);
        int snum=0,dnum=s[a].size();
        for(auto it=s[b].begin();it!=s[b].end();it++){
            if(s[a].find(*it)!=s[a].end())snum++;
            else dnum++;
        }
        double p=1.0*snum/dnum*100;
        printf("%.1f%%\n",p);   
    }
}