【PAT】1063. Set Similarity (25) 待改进

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%

---

分析：两个集合共有的元素个数 / 两个集合合并后的元素个数.

有一组超时，待改进：

#include<iostream>
#include<set>
#include<vector>
using namespace std;
int main()
{
	int N;	
	int i,count,j,t;
	cin>>N;
	vector< set<int> > vec(N);
	for(i=0; i<N; i++){
		scanf("%d",&count);
		for(j=0; j<count; j++){
			scanf("%d",&t);
			vec[i].insert(t);
		}
	}
	int request,a,b;
	scanf("%d",&request);
	for(i=0; i<request; i++){
		scanf("%d %d",&a,&b);
		set<int> temp_set;
		set<int>::iterator it;
		for(it = vec[a-1].begin(); it != vec[a-1].end(); it++){
			temp_set.insert( *it );
		}
		for(it = vec[b-1].begin(); it != vec[b-1].end(); it++){
			temp_set.insert( *it );
		}
		
		printf( "%.1f",(float)( vec[a-1].size()+vec[b-1].size()-temp_set.size())*100/(float)(temp_set.size()) );
		cout<<"%"<<endl;
	}
	return 0;
}