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 ${\mathrm{f1,f2,\dots ,fn.\; Any\; file\; name\; contains\; between\; 1\; and\; 8}}^{}$

characters, inclusive. All file names are unique.

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

${\mathrm{.\; For\; each\; query\; s\; it\; should\; count\; number\; of\; files\; containing\; s\; as\; a\; substring\; (i.e.\; some\; continuous\; segment\; of\; characters\; in\; a\; file\; name\; equals\; s}}^{}$

) 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 $n$

${\mathrm{(1\le n\le 10000}}^{}$

) — the total number of files.

The following $n$

${\mathrm{lines\; contain\; file\; names,\; one\; per\; line.\; The\; i-th\; line\; contains\; fi\; —\; the\; name\; of\; the\; i-th\; file.\; Each\; file\; name\; contains\; between\; 1\; and\; 8}}^{}$

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 $q$

${\mathrm{(1\le q\le 50000}}^{}$

) — the total number of queries.

The following $q$

${\mathrm{lines\; contain\; queries\; s1,s2,\dots ,sq,\; one\; per\; line.\; Each\; sj\; has\; length\; between\; 1\; and\; 8}}^{}$

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

Output

Print $q$

${\mathrm{lines,\; one\; per\; query.\; The\; j-th\; line\; should\; contain\; the\; response\; on\; the\; j-th\; query\; —\; two\; values\; cj\; and\; tj}}^{}$

, where

• ${\mathrm{is\; the\; number\; of\; matched\; files\; for\; the\; j-th\; query,\; tj\; is\; the\; name\; of\; any\; file\; matched\; by\; the\; j-th\; query.\; If\; there\; is\; no\; such\; file,\; print\; a\; single\; character\; \text{'}-\text{'}\; instead.\; If\; there\; are\; multiple\; matched\; files,\; print\; any.}}^{}$

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=0;i<str.size();i++)
    {
        for(int j=1;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=0;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<<"0"<<" "<<"-"<<endl;
        }
    }
}