1063 Set Similarity （25 分）

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%
思路
　　使用set即可

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<string.h>
#include<cstdio>
#include <unordered_map>
#include<cmath>

using namespace std;

int main()
{
    int n;
    scanf("%d",&n);
    set<int>s[n+1];
    int num;
    for(int i=1;i<n+1;i++)
    {
        scanf("%d",&num);
        for(int j=0;j<num;j++)
        {
            int temp;
            scanf("%d",&temp);
            s[i].insert(temp);
        }
    }
    int k;
    scanf("%d",&k);
    for(int i=0;i<k;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        set<int> result;
        set_intersection(s[a].begin(),s[a].end(),s[b].begin(),s[b].end(),inserter(result,result.begin()));
        int nc=result.size();
        int nt=s[a].size()+s[b].size()-nc;
       // cout<<nc<<" "<<nt<<endl;
        printf("%.1f\%
",nc*100.0/nt);
    }
    return 0;
}