PAT 甲级 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%

# include 
# include 
# include 
using namespace std;

int main(){
//	freopen("C:\\1.txt", "r", stdin);
	int n, m, k, dif = 0, same = 0;
	scanf("%d", &n);
	vector > v(n);
	for(int i = 0; i < n; i++){
		scanf("%d", &m);
		int tmp;
		for(int j = 0; j < m; j++){
			scanf("%d", &tmp);
			v[i].insert(tmp);
		}
	}
	scanf("%d", &k);
	for(int i = 0; i < k; i++){
		int a, b;
		scanf("%d %d", &a, &b);
		for(set::iterator it = v[a-1].begin(); it != v[a-1].end(); it++){
			set::iterator th = v[b-1].find(*it);
			if(*th < v[b-1].size()) same++;
			else if(*it == *th) same++;
		}
//		cout << same << endl;
		dif = v[a-1].size() + v[b-1].size() - same;
		double f = 100 * (double) same / dif;
		printf("%.1lf%\n", f);
		same = dif = 0;
	}
	return 0;
}