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
Copy
30 ariksihsidlihcdnaehsetahgnisol 10 Kira hates is he losing death childish L and Noteoutput
Copy
Kira is childish and he hates losinginput
Copy
12 iherehtolleh 5 HI Ho there HeLLo hellooutput
Copy
HI there HeLLoNote
In sample case 2 there may be multiple accepted outputs, "HI there HeLLo" and "HI there hello" you may output any of them.
题目大意：给你一个字符串s，它是由接下来输入的m个字符串组成的，但是是由那m个字符串都转换小写字母，并且每一个都颠倒顺表拼接成的，让你求出按照那m个字符串的形式还原出原来的字符串。具体看样例应该可以明白。
解题思路：把那m个字符串用字典树存起来，不过存的时候倒着存，因为给出的s串是由m个串颠倒组成的。然后对于字符串s从头开始在字典树上作dfs搜索就行了，每次搜到一个单词，用数组把下表保存下来最后输出即可。
/*
@Author: Top_Spirit
@Language: C++
*/
#include <bits/stdc++.h>
using namespace std ;
typedef unsigned long long ull ;
typedef long long ll ;
const int Maxn = 1e5 +10 ;
const int Max = 26;
struct Tire{
Tire *Next[Max] ;
int index ;
Tire(){
memset(Next, NULL, sizeof(Next)) ;
index = -1 ;
}
};
int n, m, flag ;
string s1 ,s2[Maxn] ;
Tire *root ;
int indexing[Maxn], re[Maxn] ;
void insert (string s, int k){
Tire *p = root ;
int len = s.size() ;
transform(s.begin(), s.end(), s.begin(), ::tolower) ;
// cout << s << endl ;
for (int i = len - 1; i >= 0; i--){
int id = s[i] - 'a' ;
if (p -> Next[id] == NULL) p -> Next[id] = new Tire() ;
p = p -> Next[id] ;
if (i == 0) p -> index = k ;
}
}
void Dfs (int key, int len, int sub){
char s[Maxn] ;
if (flag) return ;
if (key == len && !flag){
for (int i = 0; ;i++) {
if (indexing[i]) re[i] = indexing[i] ;
else break ;
}
flag = 1 ;
return ;
}
Tire *q ;
for (int i = key; i < len; i++){
if (flag) return ;
s[i - key] = s1[i] ;
if (i == key) q = root ;
int id = s[i - key] - 'a' ;
Tire *tmp = q -> Next[id] ;
if (tmp == NULL) return ;
if (tmp -> index != -1){
indexing[sub] = tmp -> index ;
Dfs(i + 1, len, sub + 1) ;
indexing[sub] = 0 ;
}
q = tmp ;
}
}
int main (){
root = new Tire() ;
cin >> n >> s1 >> m ;
for (int i = 1; i <= m; i++){
cin >> s2[i] ;
insert(s2[i], i) ;
}
Dfs(0, n, 0) ;
int up = -1 ;
for (int i = 0; ; i++){
if (!re[i]) {
up = i ;
break ;
}
}
// cout << up << endl ;
cout << s2[re[0]] ;
for (int i = 1; i <= up; i++){
cout << " " << s2[re[i]] ;
}
cout << endl ;
return 0 ;
}
古今多少事
都付笑谈中