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.
解答：这道题我刚开始用set<int> a[n]来保存数据，排序，再用一个set来求两个集合的交集，但出现超时的情况。苦思冥想，不得已借鉴了其他博主的想法，即用set_union和set_intersection来得到两个集合的并集和交集，顺利AC。但我还是不了解为什么这两个函数的效率如此之高。。先记住，之后慢慢探索吧。
AC代码如下：
#include<iostream>
#include<vector>
#include<algorithm>
#include<set>using namespace std;int main()
{int n;scanf("%d", &n);set<int> sets[n+5];for(int i = 1; i <= n; ++i){int m;scanf("%d", &m);for(int j = 0; j < m; ++j){int val;scanf("%d", &val);sets[i].insert(val);}}//进行查询处理 int k;scanf("%d", &k); for(int i = 0; i < k; ++i){int setn1, setn2;vector<int> nc_vec, nt_vec;scanf("%d %d", &setn1, &setn2);set_union(sets[setn1].begin(), sets[setn1].end(), sets[setn2].begin(), sets[setn2].end(), back_inserter(nt_vec));set_intersection(sets[setn1].begin(), sets[setn1].end(), sets[setn2].begin(), sets[setn2].end(), back_inserter(nc_vec));int nc = nc_vec.size(), nt = nt_vec.size();double rate = 100.0 * nc / nt;printf("%.1f%\n", rate);}return 0;
}