a1063 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%

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题意：

输入分两部分：

1.多组数据，每组数据分别输入：该组的数据数量，每个数据

2.多组数据，分别输入该组要比较的两个set

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思路：

要求的是两个set相同的数字所占比例，设两个set为s1，s2，两个set合并后为s3，那么这个比例可以表示为：(s1.size()+s2.size()-s3.size)/s3.size()

1.利用set<int> s[55]将每组数据存入数组

2.创建中间用于比较的set t0.

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注意：

1.每次t0都要clear一下

2.注意最后保留一位小数且使用百分数形式表示

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ac代码：

#include<iostream>
#include<set>
#include<iomanip>
using namespace std;
int main(){
	int n,a,b,x,y,s1,s2,t0;
	cin>>n;
	set<int> s[55];
	set<int> t;
	for(int i=1;i<=n;i++){
		cin>>a;
		for(int j=0;j<a;j++){
			cin>>b;
			s[i].insert(b);
		}
	}
	cin>>a;
	for(int i=0;i<a;i++){
		cin>>x>>y;
		s1=s[x].size();
		s2=s[y].size();
		t.insert(s[x].begin(),s[x].end());
		t.insert(s[y].begin(),s[y].end());
		t0=t.size();
		cout<<setiosflags(ios::fixed)<<setprecision(1)<<(double)(s1+s2-t0)*100/t0<<"%"<<endl;
        t.clear();
	}
}