BerOS File Suggestion(stl-map应用）

Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.

There are n

files on hard drive and their names are f1,f2,…,fn. Any file name contains between 1 and 8

characters, inclusive. All file names are unique.

The file suggestion feature handles queries, each represented by a string s

. For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s

) and suggest any such file name.

For example, if file names are "read.me", "hosts", "ops", and "beros.18", and the query is "os", the number of matched files is 2

(two file names contain "os" as a substring) and suggested file name can be either "hosts" or "beros.18".

Input

The first line of the input contains integer n

(1≤n≤10000

) — the total number of files.

The following n

lines contain file names, one per line. The i-th line contains fi — the name of the i-th file. Each file name contains between 1 and 8

characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters ('.'). Any sequence of valid characters can be a file name (for example, in BerOS ".", ".." and "..." are valid file names). All file names are unique.

The following line contains integer q

(1≤q≤50000

) — the total number of queries.

The following q

lines contain queries s1,s2,…,sq, one per line. Each sj has length between 1 and 8

characters, inclusive. It contains only lowercase Latin letters, digits and dot characters ('.').

Output

Print q

lines, one per query. The j-th line should contain the response on the j-th query — two values cj and tj

, where

• cjis the number of matched files for the j-th query, tj is the name of any file matched by the j-th query. If there is no such file, print a single character '-' instead. If there are multiple matched files, print any.

4
test
contests
test.
.test
6
ts
.
st.
.test
contes.
st

Output

Copy

1 contests
2 .test
1 test.
1 .test
0 -
4 test.

题意：有n个字符串，q次查询，输出与之匹配字符串的个数，并输出任意一个字符串;

题解：由于每个字符串的长度<=8,所以我们可以把该字符串的字串全部存进map，并记录该字符串的位置;

直接输出就可以啦;

 #include<cstdio>
 #include<cstring>
 #include<stack>
 #include<queue>
 #include<algorithm>
 #include<iostream>
 #include<map>
 #include<vector>
 #define PI acos(-1.0)
 using namespace std;
 typedef long long ll;
 int m,n;
 map<string,int>::iterator it;
 map<string,int>mp;//存储所有字符串的字串
 map<string,int>pos;//存储字符串的位置
 void Substr(string str,int num)
 {
     string arr;
     map<string,int>mp1;
     for(int i=;i<str.size();i++)
     {
         for(int j=;j<=str.size();j++)
         {
             arr=str.substr(i,j);
             if(!mp1.count(arr))
             {
                 mp1[arr]++;
                 pos[arr]=num;
             }
         }
     }
     for(it=mp1.begin();it!=mp1.end();it++)
     {
         mp[it->first]++;
         //cout<<it->first<<endl;
     }
 }
 int main()
 {
     string s,str;
     cin>>m;
     map<int,string>kp;//记录原串
     for(int i=;i<m;i++)
     {
         cin>>str;
         kp[i]=str;
         Substr(str,i);//求子串
     }
     cin>>n;
     while(n--)
     {
         cin>>s;
         if(mp.count(s))
         {
             cout<<mp[s]<<" "<<kp[pos[s]]<<endl;
         }
         else
         {
         cout<<""<<" "<<"-"<<endl;
         }
     }
 }