BerOS File Suggestion(字符串匹配map)

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.

MAP是个好东西
题意：
给你n个字符串，再给你q次查询，问你每一次查询的字符串是n个字符串中多少个字符串的子串，并随机输出其中一个原串

    分析：将给出的n个字符串的所有子串全部存起来，长度为1的子串，长度为2....长度为size的子串，再将查询的字符串与所有的子串比对。再开一个map，用来记录该子串所在的原串。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<map>
 5 #include<iostream>
 6 using namespace std;
 7 map<string,int> strr;
 8 map<string,string> sstr;
 9 int main()
10 {
11     int n;
12     while(~scanf("%d",&n)) {
13         while(n--) {
14             string a;
15             cin>>a;
16             map<string,int> outrepeat;//去重，防止一个字符串中，同样的子串出现不止一次
17             for(int i=0;i<a.size();i++) {
18                 for(int j=1;j<=a.size();j++) {
19                     string b;
20                     b=a.substr(i,j);
21                     if(outrepeat[b]==0) {//去重
22                         strr[b]++;//该子串在n个原串中出现的次数
23                         sstr[b]=a;//记录该子串所属的原串
24                         outrepeat[b]=1;
25                     }
26                 }
27             }
28         }
29        int m;
30        scanf("%d",&m);
31        while(m--) {
32         string c;
33         cin>>c;
34         if(strr[c]>0)
35         cout<<strr[c]<<" "<<sstr[c]<<endl;
36         else
37         cout<<"0 -"<<endl;
38        }
39     }
40     return 0;
41 }