codeforces--510C--Fox And Names

Fox And Names

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order.

After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted inlexicographical order in normal sense. But it was always
true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical!

She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order.
If so, you should find out any such order.

Lexicographical order is defined in following way. When we compare s and t,
first we find the leftmost position with differing characters: si ≠ ti.
If there is no such position (i. e. s is a prefix of t or
vice versa) the shortest string is less. Otherwise, we compare characters si and tiaccording
to their order in alphabet.

Input

The first line contains an integer n (1 ≤ n ≤ 100):
number of names.

Each of the following n lines contain one string namei (1 ≤ |namei| ≤ 100),
the i-th name. Each name contains only lowercase Latin letters. All names are different.

Output

If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on).

Otherwise output a single word "Impossible" (without quotes).

Sample test(s)

input

3
rivest
shamir
adleman

output

bcdefghijklmnopqrsatuvwxyz

input

10
tourist
petr
wjmzbmr
yeputons
vepifanov
scottwu
oooooooooooooooo
subscriber
rowdark
tankengineer

output

Impossible

input

10
petr
egor
endagorion
feferivan
ilovetanyaromanova
kostka
dmitriyh
maratsnowbear
bredorjaguarturnik
cgyforever

output

aghjlnopefikdmbcqrstuvwxyz

input

7
car
care
careful
carefully
becarefuldontforgetsomething
otherwiseyouwillbehacked
goodluck

output

acbdefhijklmnogpqrstuvwxyz

 

 

给出按字典序排序好的字符串，问是否能写出字典序来，或者不能写出。

注意判断各种情况，两个字符串比较的时候，从左向右开始比较，当出现不同的字母后，不再向后比较。一个是另一个的前缀，那么长的在后面。

当给出的顺序出现环的时候，或者有串的前缀在串的后面时，出现错误。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std ;
struct node
{
int v , next ;
} p[10000000];
int head[50] , cnt ;
int in[50] , vis[110] ;
char str[110][110] ;
char s[50] , num ;
queue <int> que ;
void add(int u,int v)
{
p[cnt].v = v ;
p[cnt].next = head[u] ;
head[u] = cnt++ ;
return ;
}
int main()
{
int n , i , j , m = 0 ,x , k , l ;
memset(head,-1,sizeof(head)) ;
memset(in,0,sizeof(in)) ;
memset(vis,0,sizeof(vis)) ;
cnt = 0 ;
scanf("%d", &n) ;
for(i = 0 ; i < n ; i++)
{
scanf("%s", str[i] ) ;
x = strlen(str[i]) ;
m = max(m,x) ;
}
int flag = 0 ;
for(i = 0 ; i < m ; i++)
{
for(j = 0 ; j < n ; j++)
{
for(k = j-1 ; k >= 0 ; k--)
{
if( !(str[k][i] >= 'a' && str[k][i] <= 'z') ) continue ;
for(l = 0 ; l < i ; l++)
if( str[j][l] != str[k][l] )
break ;
if( l < i ) continue ;
if( str[j][i] == '\0' && str[k][i] != '\0' )
flag = 1 ;
if( flag )
{
flag = 1 ;
break ;
}
if( str[j][i] != str[k][i] )
{
add(str[k][i]-'a',str[j][i]-'a') ;
in[ str[j][i]-'a' ]++ ;
}
}
if( flag )
break ;
}
if( flag )
break ;
}
if( flag )
{
printf("Impossible\n") ;
return 0;
}
num = 0 ;
while( !que.empty() ) que.pop() ;
for(i = 0 ; i < 26 ; i++)
if( in[i] == 0 )
{
que.push(i) ;
s[num++] = i + 'a' ;
}
int u , v ;
while( !que.empty() )
{
u = que.front() ;
que.pop() ;
for(i = head[u] ; i != -1 ;i = p[i].next)
{
v = p[i].v ;
in[v]-- ;
if( in[v] == 0 )
{
que.push(v) ;
s[num++] = v + 'a' ;
}
}
}
s[num] = '\0' ;
if( num == 26 )
{
printf("%s", s);
}
else
printf("Impossible\n") ;
return 0 ;
}