6.2 PAT A1063 Set Similarity (25分)（set用法）(自动有序并去重）

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%

题意为求给定集合的并集和交集，即公共元素的比例。注意要去重，并按查询输入求出对应两组的相同元素比例。可以利用set来求解。

参考代码：

#include <cstdio>
#include <set>
using namespace std;
const int N=52;
set<int> st[N];
void compare(int x,int y)
{
	int total =st[y].size(); //注意是y的个数 
	int same=0;
	for(set<int>::iterator it=st[x].begin();it!=st[x].end();++it) //迭代器写法 
	{
		if(st[y].find(*it)!=st[y].end()) //利用查找函数查询是否有相同元素 
			same++; 
		else
			total++;  //不同元素个数 
	}
	printf("%.1f%%\n",same*100.0/total); //输出比例 
 } 
int main()
{
	int n,k,m,v,s1,s2;
	while(scanf("%d",&n)!=EOF)
	{
		
		for(int i=1;i<=n;++i)
		{
			scanf("%d",&m);
			for(int j=0;j<m;++j)
			{
				scanf("%d",&v);
				st[i].insert(v);
			}
		}
		scanf("%d",&k); //输入查询个数 
		for(int i=0;i<k;++i)
		{
			scanf("%d%d",&s1,&s2); //查询组 
			compare(s1,s2);
		}
	}
	return 0;
 }