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%

题目大意：

给定N个集合，给出的集合中可能含有相同的值。然后要求M个查询，每个查询给出两个集合的编号x和y，要求集合x和集合y的相同元素率：。

解题思路：

这里我们可以借助N个STL的set容器来保存这N个set，当判断x和y两个set的相同元素率时，我们只需要遍历这两个set统计其中的重复元素的个数即可。 

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

set<int> s[60];

double fun(int x, int y)
{
	int cnt = 0;
	for(set<int>::iterator it = s[x].begin();it!=s[x].end();it++)
	{
		if(s[y].find(*it)!=s[y].end())
			cnt++;
	}
	double res = cnt*100.0 / (s[x].size() + s[y].size() - cnt);
	return res;
}

int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		int m;
		scanf("%d", &m);
		for (int j = 0; j < m; j++)
		{
			int num;
			scanf("%d", &num);
			s[i].insert(num);
		}
	}
	int k;
	scanf("%d", &k);
	for (int i = 0; i < k; i++)
	{
		int x, y;
		scanf("%d %d", &x, &y);
		printf("%.1lf%%\n", fun(x, y));
	}
	return 0;
}