本文主要是介绍CodeForces 633C Spy Syndrome 2，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：http://codeforces.com/problemset/problem/633/C
Spy Syndrome 2
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 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 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
30
ariksihsidlihcdnaehsetahgnisol
10
Kira
hates
is
he
losing
death
childish
L
and
Note
Kira is childish and he hates losing
12
iherehtolleh
5
HI
Ho
there
HeLLo
hello
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.
思路：其实还是挺好想的，关键就是看你敢不敢这样写。将所有的模式串倒着来建立一棵字典树，然后从文本串第一个位置开始深搜，如果能够到达一个单词节点，继续深搜，直到文本串所有位置全部匹配，否则回溯到原来位置继续匹配。如果到达不了单词节点，说明没有能够匹配的模式串，直接退出。详见代码。注意：这么大的数据，DFS居然不超时，很是疑惑啊！若有大神路过，望指点迷津，先说声谢谢！
附上AC代码：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
string text, p[maxn];
int trie[maxn<<4][26];
int ans[maxn>>3], val[maxn<<4];
int n, m, cnt, total;
bool ok;void _insert(int x){int pos = 0;for (int i=p[x].size()-1; i>=0; --i){int id = tolower(p[x][i])-'a';if (!trie[pos][id])trie[pos][id] = cnt++;pos = trie[pos][id];}val[pos] = x;
}void dfs(int len){if (len == n-1){for (int i=0; i<total; ++i)cout << p[ans[i]] << (i!=total-1 ? ' ' : '\n');ok = true;return ;}int pos = 0;for (int i=len+1; !ok&&i<n; ++i){int id = text[i]-'a';pos = trie[pos][id];if (!pos)break;if (val[pos]){ans[total++] = val[pos];dfs(len+p[val[pos]].size());--total;}}
}int main(){ios::sync_with_stdio(false);cin.tie(0);cin >> n >> text >> m;cnt = 1;for (int i=1; i<=m; ++i){cin >> p[i];_insert(i);}dfs(-1);return 0;
}
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