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 (≤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%

题目描述：求两个集合中，相同元素在两个集合中占比为多少。

输入数据：给n组集合，每组集合有m个元素，题目给出k组查询，每组询问两集合中相同元素的占比。

分析输入样例

查询1、2集合：相同数2（87 101）/总数4（87 101 99 87 5）=0.5
查询1、3集合：相同数2（99 101）/总数6（99 87 101 18 5 135）=0.3

注意最后一个测试点容易超时，使用set的find()。

#include<bits/stdc++.h>
using namespace std;
vector< set<int> > v;
int main()
{
	int n, m, k, g;
	scanf("%d", &n);
	v.resize(n+1);
	for(int i=1; i<=n; i++){
		scanf("%d", &m);
		for(int j=0; j<m; j++){
			scanf("%d", &g);
			v[i].insert(g);
		}
	}
	int count=0, all_num=0, a, b;
	scanf("%d", &k);
	set<int>::iterator it;
	for(int i=0; i<k; i++){
		scanf("%d %d", &a, &b);
		all_num = v[a].size()+v[b].size();
		count = 0;
		for(it=v[a].begin(); it!=v[a].end(); it++){
			if(v[b].find(*it)!=v[b].end()) count++;//在集合b中可以找到 
		}
		printf("%.1f%\n", count*100.0/(all_num-count));//共同的元素出现了两次 
	}
	return 0;
}