Codeforces 633C Spy Syndrome 2 【字典树 + DFS】

C. Spy Syndrome 2

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and ancient as Caesar cipher. After many weeks of observation of Siddhant’s sentences, Yash determined
a new cipher technique.

For a given sentence, the cipher is processed as:

Convert all letters of the sentence to lowercase.

Reverse each of the words of the sentence individually.

Remove all the spaces in the sentence.

For example, when this cipher is applied to the sentence

Kira is childish and he hates losing
the resulting string is

ariksihsidlihcdnaehsetahgnisol

Now Yash is given some ciphered string and a list of words. Help him to find out any original sentence composed using only words from the list. Note, that any of the given words could be used in the sentence multiple times.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 10 000) —
the length of the ciphered text. The second line consists of nlowercase English letters — the ciphered text t.

The third line contains a single integer m (1 ≤ m ≤ 100 000) —
the number of words which will be considered while deciphering the text. Each of the next m lines contains a non-empty word wi (|wi| ≤ 1 000)
consisting of uppercase and lowercase English letters only. It's guaranteed that the total length of all words doesn't exceed 1 000 000.

Output

Print one line — the original sentence. It is guaranteed that at least one solution exists. If there are multiple solutions, you may output any of those.

Examples

input

30
ariksihsidlihcdnaehsetahgnisol
10
Kira
hates
is
he
losing
death
childish
L
and
Note

output

Kira is childish and he hates losing

input

12
iherehtolleh
5
HI
Ho
there
HeLLo
hello

output

HI there HeLLo

Note

In sample case 2 there may be multiple accepted outputs, "HI there HeLLo" and "HI there hello" you may output any of them.

题意：给定一个只有小写字母组成的目标串和m个模式串(里面可能有大写字母)，记目标串反过来后的串为S，让你从m个模式串中选出若干个组成S串(不区分大小写)。输出任意一种方案。

思路：建好trie，就是DFS了。不知为什么倒着DFS会T，只好记忆化一下。

 

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <stack>
#include <algorithm>
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
const int MAXN = 1e6+10;
int next[MAXN][30], word[MAXN], End[MAXN];
int root, L;
int newnode() {
for(int i = 0; i <= 25; i++)
next[L][i] = -1;
word[L] = 0; End[L] = -1; L++;
return L-1;
}
void init() {L = 0; root = newnode();}
void Insert(char *s, int id)
{
int u = root, v;
for(int i = 0; s[i]; i++)
{
v = s[i] - 'a';
if(s[i] >= 'A' && s[i] <= 'Z')
v = s[i] - 'A';
if(next[u][v] == -1)
next[u][v] = newnode();
u = next[u][v];
word[u]++;
}
End[u] = id;
}
const int MAXM = 1e4+10;
char str[MAXM]; int n;
const int N = 1e5+10;
char ss
[1010];
int cnt;
int mark[MAXM];
bool DFS(int pos)
{
if(pos == -1) return true;
if(mark[pos] != -1) return mark[pos];
int u = root, v;
for(int i = pos; i >= 0; i--)
{
int v = str[i] - 'a';
if(next[u][v] == -1) break;
u = next[u][v];
int id = End[u];
if(id != -1 && DFS(i - 1)) {
if(cnt) cout << " ";
cout << ss[id]; cnt++;
return mark[pos] = true;
}
}
return mark[pos] = false;
}
int main()
{
cin >> n; cin >> str;
int m; cin >> m; init();
for(int i = 0; i < m; i++)
{
cin >> ss[i];
Insert(ss[i], i);
}
CLR(mark, -1); cnt = 0; DFS(n-1);
return 0;
}