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
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.
題意：
給你一個加密後的字符串，再給你若干個單詞。
加密規則：
1、把這句話中的所有字母變成小寫字母
2、把這句話中的所有單詞反過來
3、把這句話中的所有空格去掉
現在給你加密後的字符串和字典，問你加密前的句子是什麼，輸出任意一組合法答案。
思路：剛開始以爲是個簡單的字典樹，可是怎麼調都不對。後來一想，要想合法答案存在，必定是從最後一個單詞確定開始，依次找前面的單詞的。因此用一個dp數組表示下標i之前的串是哪個單詞（i一定是單詞的結束位置）。在字典樹中dp完以後再從n開始倒着往前找單詞即可。
注意細節。
代碼：
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define p pair<int,int>
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];//注意數組大小
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<<u<<endl;
}
void find(char *s,int id)
{
int u=0;
for(int i=id;i<n;i++)
{
u=ch[u][s[i]-'a'];
if(!u) return;
if(val[u]&&!d[i+1])
{
d[i+1]=val[u];
}
}
}
}T;
int main()
{
T.init();
scanf("%d",&n);
scanf("%s",ss);
scanf("%d",&m);
for(int i=1;i<=m;i++)
{
scanf("%s",s);
str[i]=(string)s;
len[i]=strlen(s);
T.add(s,i);
}
d[0]=1;
for(int i=0;i<n;i++)
{
if(!d[i]) continue;
T.find(ss,i);
}
vector<int>ans;
int u=n;
while(u)
{
ans.push_back(d[u]);
//cout<<d[u]<<" "<<len[d[u]]<<endl;
u-=len[d[u]];
//cout<<u<<endl;
}
for(int i=ans.size()-1;i>=0;i--)
{
printf("%s%c",str[ans[i]].c_str(),i==0?'\n':' ');
}
return 0;
}