CodeForces - 633C - Spy Syndrome 2（字典树+dp）

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:

1. Convert all letters of the sentence to lowercase.
2. Reverse each of the words of the sentence individually.
3. 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

Copy

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

output

Copy

Kira is childish and he hates losing 

input

Copy

12
iherehtolleh
5
HI
Ho
there
HeLLo
hello

output

Copy

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：

给你一个加密后的字符串，再给你若干个单词。

加密规则：

1、把这句话中的所有字母变成小写字母

2、把这句话中的所有单词反过来

3、把这句话中的所有空格去掉

现在给你加密后的字符串和字典，问你加密前的句子是什么，输出任意一组合法答案。

思路：刚开始以为是个简单的字典树，可是怎么调都不对。后来一想，要想合法答案存在，必定是从最后一个单词确定开始，依次找前面的单词的。因此用一个dp数组表示下标i之前的串是哪个单词（i一定是单词的结束位置）。在字典树中dp完以后再从n开始倒着往前找单词即可。

注意细节。

代码：

#include
#define ll long long
#define inf 0x3f3f3f3f
#define p pair
using namespace std;
const int maxn=200010;
const ll mo=998244353;
int n,m,k;
int d[maxn],len[maxn];
int tmp,ans;
string str[maxn];
char s[1010],ss[maxn];
struct Trie
{
    int ch[1000010][26],sz;
    int val[1000010];//注意数组大小
    void init()
    {
        memset(ch[0],0,sizeof(ch[0]));
        sz=0;
        val[sz]=0;
    }
    void add(char *s,int idd)
    {
        int u=0,id;
        //printf("%s\n",s);
        for(int i=len[idd]-1;i>=0;i--)
        {
            if(s[i]<='Z') id=s[i]-'A';
            else id=s[i]-'a';
            if(!ch[u][id])
            {
                ch[u][id]=++sz;
                val[sz]=0;
                memset(ch[sz],0,sizeof(ch[sz]));
            }
            u=ch[u][id];
        }
        val[u]=idd;
        //if(s[0]=='h'&&s[1]=='e') cout<ans;
    int u=n;
    while(u)
    {
        ans.push_back(d[u]);
        //cout<=0;i--)
    {
        printf("%s%c",str[ans[i]].c_str(),i==0?'\n':' ');
    }
    return 0;
}