1063 Set Similarity (25 point(s))

1063 Set Similarity (25 point(s))

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%

考点：STL set的使用。相似度定义：交集元素个数和并集元素个数的比例。

参考链接：https://blog.youkuaiyun.com/changjiale110/article/details/79108447

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

int main(void){
	int N,K,M,a,b;
	cin>>N;
	set<int> sets[N+1];
	for(int i=1;i<=N;i++){
		cin>>M;
		for(int j=0;j<M;j++){
			cin>>a;
			sets[i].insert(a);
		}
	}
	cin>>K;
	set<int>::iterator it; 
	while(K--){
		cin>>a>>b;
		int cnt=0;//交集元素
		for(it=sets[a].begin();it!=sets[a].end();it++){
			if(sets[b].find(*it)!=sets[b].end()) cnt++;
		} 
		double ans = 100.00*(double)cnt/double(sets[a].size()+sets[b].size()-cnt);
		printf("%.1f%%\n",ans);
	}
	
	return 0;
}