CodeForces - 633C(字典树+dfs回溯)

缺乏、安全感 2022-05-28 13:05 138阅读 0赞

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 *n*lowercase 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 *w**i* (|*w**i*| ≤ 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

Kira is childish and he hates losing

input

Copy

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.

#include<cstdio>
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    #include<map>
    using namespace std;
    typedef long long ll;
    const int maxn=1e5+10;
    int n, m;
    char s[maxn], word[maxn][1005];
    int trie[1000005][27], tot=0, mark[1000005], ans[maxn], cnt=0;//开数组一定要开的刚好，一开始数组开小了，wa了好几次，字典树一般开1e6*27的空间
    
    void insert(int len, int id){//字典树模板
    	
    	int now=0;
    	for(int i=len-1; i>=0; i--)
    	{
    		int to;
    		if(word[id][i]>'Z') to=word[id][i]-'a';
    		else to=word[id][i]-'A';
    		
    		if(!trie[now][to])
    			trie[now][to]=++tot;
    			
    		now=trie[now][to];
    	}
    	mark[now]=id;
    }
    
    bool solve(int pos){
    	if(pos==n)
    	{
    		for(int i=0; i<cnt; i++)
    			printf("%s%s", word[ans[i]], i==cnt-1 ? "\n":" ");
    		return true;
    	}
    	int now=0;
    	for(int i=pos; i<n; i++)
    	{
    		int to=s[i]-'a';
    		
    		if(trie[now][to]==0) break;
    		now=trie[now][to];
    		if(mark[now])
    		{
    			ans[cnt++]=mark[now];
    			if(solve(i+1)) return true;
    			cnt--;
    		}	
    	}
    	return false;
    }
    
    int main(){
    	scanf("%d%s", &n, s);
    	scanf("%d", &m);
    	
    	for(int i=1; i<=m; i++)
    	{
    		scanf("%s", word[i]);
    		int len=strlen(word[i]);
    		insert(len, i);
    		
    	}
    	
    	solve(0);
    	
    	return 0;
    }