【PAT1063】Set Similarity 求两集合的交集、并集
1063. Set Similarity (25)
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 Nt 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 (<=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 3Sample Output:
50.0% 33.3%
分析：使用set可以在insert时直接过滤掉重复的数值；使用set_intersection函数可以求两集合交集，set_union函数求并集。
代码：
#include <iostream>
#include <fstream>
#include <algorithm> //set_intersection 函数
#include <iomanip>
#include <iterator> //inserter函数
#include <set>
using namespace std;
#include <stdio.h>
//此代码使用前，需删除下面两行+后面的system("PAUSE")
ifstream fin("in.txt");
#define cin fin
int main()
{
int n,m,in;
cin>>n;
set<int> *s = new set<int>[n];
int i,j;
for(i=0;i<n;i++)
{
cin>>m;
for(j=0;j<m;j++)
{
cin>>in;
s[i].insert(in);
}
}
set<int> res;
int k,a,b,nc,nt;
cin>>k;
for(i=0;i<k;i++)
{
cin>>a>>b;
set_intersection(s[a-1].begin(),s[a-1].end(),s[b-1].begin(),s[b-1].end(),inserter(res,res.begin())); //求集合交集
nc = res.size();
nt = s[a-1].size()+s[b-1].size()-nc;
res.clear();
//set_union(s[a-1].begin(),s[a-1].end(),s[b-1].begin(),s[b-1].end(),inserter(res,res.begin())); //求集合并集，可通过数值计算得出
//nt = res.size();
//res.clear();
cout<<setiosflags(ios::fixed)<<setprecision(1)<<(double)nc*100/nt<<"%"<<endl;
}
system("PAUSE");
return 0;
}