Codeforces 510C （拓扑排序）

原题：http://codeforces.com/problemset/problem/510/C

C. 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 in lexicographical 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).

Examples

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

******************************************************************************************************************

题解：

　　　　给你n个字符串（全小写），让你按照输入的顺序来个字母表排列，（就是改变字母表的某些字母位置，使得你的输入是按照新的字典序）

　　　　如果前一个是后一个的子串，那么前一个一定小于后一个所以可以跳过。反之，如果后一个是前一个的子串，无论字典序怎么改都无法成立就输出"Impossible"。

　　　　新的字典序就是按照2个字符串的不同首字母，构建的有向图。

　　　　在使用拓扑排序就可以了。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <vector>
 #include <map>
 #include <set>
 #include <queue>
 #include <sstream>
 #include <algorithm>
 using namespace std;
 #define pb push_back
 #define mp make_pair
 #define mset(a, b)  memset((a), (b), sizeof(a))
 typedef long long LL;
 const int inf = 0x3f3f3f3f;
 const int maxn = +;
 string s[maxn];
 int gap[][];
 int c[], topo[], t;
 bool dfs(int u)
 {
     c[u] = -;
     for(int  v = ; v<; v++)   if(gap[u][v]){
         if(c[v] <  )   return false;
         else if(!c[v] && !dfs(v))   return false;
     }
     c[u] = ; topo[--t] = u;
     return true;
 }
 bool toposort()
 {
     t = ;
     mset(c, );
     for(int i=;i<;i++)   if(!c[i])
         if(!dfs(i)) return false;
     return true;
 }
 int main()
 {
     int n;
     cin >> n;
     for(int i=;i<n;i++)    cin >> s[i];
     for(int i=;i<n-;i++){
         int len1 = s[i].size();
         int len2 = s[i+].size();
         int p = ;
         while(p<len1 && p<len2 && s[i][p]==s[i+][p])   p++;
         if(p == len1 && len1 < len2)    continue;
         if(p == len2 && len2 < len1)    {cout<<"Impossible"<<endl;return ;}
         if(gap[s[i][p]-'a'][s[i+][p] -'a'] == )     continue;
         gap[s[i][p]-'a'][s[i+][p]-'a']=;
     }
     if(toposort()){
         for(int i=;i<;i++)
             printf("%c", topo[i]+'a');
     }
     else
         cout <<"Impossible"<< endl;
     return ;
 }