codeforces Round #451 (Div. 2) Phone Numbers

思路：直接暴力就好了,写这道题主要还是熟悉一下对stl的用法。
每次插入map去重，然后每次双重循环判断是否是某个数的字串。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <set>
#include <map>
#include <vector>
using namespace std;
typedef long long LL;
map<string, set<string> > mp;
map<string, set<string> > ans;
bool judge(string s1, string s2)
{
    if(s1.size()>s2.size())
        return false;
    for(int i=0;i<s1.size();++i)
    {
        if(s1[i]!=s2[s2.size()-s1.size()+i])
            return false;
    }
    return true;
}
/*
bool judge(string s1, string s2)
{
    if(s1.size()>=s2.size())
        return false;
    int x=s2.find(s1,s2.size()-s1.size());
    //cout<<s1<<" "<<s2<<" "<<x<<endl;
    if(x==s2.size()-s1.size())
        return true;
    else
        return false;
}
*/
int main()
{
    int n, num;
    string s, ss;
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=0;i<n;++i)
    {
        cin>>s>>num;
        for(int j=0;j<num;++j)
        {
            cin>>ss;
            mp[s].insert(ss);
        }
    }
    cout<<mp.size()<<endl;
    map<string, set<string> >::iterator it;
    set<string> ::iterator its, itss;
    for(it=mp.begin();it!=mp.end();++it)
    {
        for(its=(*it).second.begin();its!=(*it).second.end();++its)
        {
            string s1=*its;
            int flag=0;
            for(itss=(*it).second.begin();itss!=(*it).second.end();++itss)
            {
                string s2=*itss;
                if(itss==its)
                    continue;
                if(judge(s1,s2))
                {
                    flag=1;
                    break;
                    //cout<<*its<<endl;
                }
            }
            if(!flag)
            {
                ans[it->first].insert(s1);
            }
        }
    }
    for(it=ans.begin();it!=ans.end();++it)
    {
        cout<<it->first<<" "<<it->second.size();
        for(its=it->second.begin();its!=it->second.end();its++)
        {
            cout<<" "<<*its;
        }
        cout<<"\n";
    }
    return 0;
}