1063 Set Similarity （25 point(s))

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本文介绍了一种计算两个集合相似度的方法，通过计算两个集合中共同元素的数量与总元素数量的比例来衡量它们的相似性。使用C++实现了一个程序，该程序能够接收多个集合并计算任意两个集合之间的相似度。

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Given two sets of integers, the similarity of the sets is defined to be Nc/Nt×100%, where Nc is the number of distinct common numbers shared by the two sets, and Ntis 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%

题目大意：题目要求从两个集合中不重复的找到公共部分Nc和所有不重复元素Nt，然后计算Nc占Nt的比例。

分析：集合用set存储（自动过滤相同元素），遍历set内元素进行查询即可

AC代码：

#include<bits/stdc++.h>
using namespace std;
int main() {
    int n,m,k,val,s1,s2;
    cin>>n;
    vector<set<int> >v(n);
    for(int i=0;i<n;i++){
        cin>>m;
        for(int j=0;j<m;j++) {
            cin>>val;
            v[i].insert(val);
        }
    }
    cin>>k;
    for(int i=0;i<k;i++){
        cin>>s1>>s2;
        int sam=0,total=v[s2-1].size();
        for(auto it=v[s1-1].begin();it!=v[s1-1].end();it++) {
            if(v[s2-1].find(*it)!=v[s2-1].end())
                sam++;
            else   
			    total++;
        }
        printf("%.1f%%\n",(double)sam/total*100);
    }
    return 0;
}