1063. Set Similarity (25)

the problem is from PAT,which website is http://pat.zju.edu.cn/contests/pat-a-practise/1063

At first, i decide use a for loop to get the intersection, which time complexity is

O(n * log(n)) ,but it turned out that the result of one example is time running out. then i

searched the web and found there are one available in the “algorithm”file.

and the code is copied from another blog ,which url is http://blog.csdn.net/tiantangrenjian/article/details/16868399

#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>



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();

        cout<<setiosflags(ios::fixed)<<setprecision(1)<<(double)nc*100/nt<<"%"<<endl;

    }

    return 0;

}