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".
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 ('.').
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
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;
}
}
}