BerOS File Suggestion

题意：

给出几个字符串，再给出几个字符串，找这些字符串在前面已给出自符串的母串有几个，输出母串个数及任意一个母串;

题解：

直接暴力，超时；每个字符串长度最大为8 ，所以把每个字符串拆成子串，用map存起来，再输入所要查找的字符串时，直接找就行。

Description

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

• cj

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 '-' instead. If there are multiple matched files, print any.

  Sample Input

  Input

  4
  test
  contests
  test.
  .test
  6
  ts
  .
  st.
  .test
  contest.
  st
  

  Output

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

代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<map>
using namespace std;
#define maxn 100010
map<string,int>str1;
map<string,string>str2;
int main()
{
    int n,i,j,m;
    string a,b,c;
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            cin>>a;
            map<string,int>outrepeat;
            for(i=0;i<a.size();i++)
            {
                for(j=1;j<=a.size();j++)
                {
                    b=a.substr(i,j);
                    cout<<b<<endl;
                    if(outrepeat[b]==0)
                    {
                        str1[b]++;
                        str2[b]=a;
                        outrepeat[b]=1;
                    }
                }
            }
        }
        cin>>m;
        while(m--)
        {
            cin>>c;
            if(str1[c]>0)
                cout<<str1[c]<<" "<<str2[c]<<endl;
            else
                cout<<"0 -"<<endl;
        }
    }
    return 0;
}