1063. Set Similarity (25)

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 (<=10⁴) and followed by M integers in the range [0, 10⁹]. 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：Nc代表的是两个集合的交集，Nt代表的是两个集合的并集。

#include<iostream>
#include<set>
#include<vector>
#include<cstdio>
using namespace std;
#define max 52
int main(){
	int n;
	scanf("%d",&n);
	vector<set<int> >v(n+1);
	int i,j;
	int key;
	int m;
	for(i=0;i<n;i++){
		scanf("%d",&m);
		while(m--){
			scanf("%d",&key);
			v[i].insert(key);
		}
	}
	int k;
	scanf("%d",&k);
	while(k--){
		int pair1;
		int pair2;
		scanf("%d%d",&pair1,&pair2);
		pair1--;
		pair2--; 
		int nt=v[pair1].size()+v[pair2].size();
		int nc=0;
		set<int>::const_iterator it;
		for(it = v[pair1].begin();it!=v[pair1].end();it++){
			if(v[pair2].count(*it)>0){
				nc++;
			}
		} 
		printf("%.1f%%\n",100*(nc/((nt-nc)*1.0)));
	}
}