Codeforces Round #290 (Div. 2) C. Fox And Names 拓扑排序

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 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 andti according 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

3rivestshamiradleman

output

bcdefghijklmnopqrsatuvwxyz

input

10touristpetrwjmzbmryeputonsvepifanovscottwuoooooooooooooooosubscriberrowdarktankengineer

output

Impossible

input

10petregorendagorionfeferivanilovetanyaromanovakostkadmitriyhmaratsnowbearbredorjaguarturnikcgyforever

output

aghjlnopefikdmbcqrstuvwxyz

input

7carcarecarefulcarefullybecarefuldontforgetsomethingotherwiseyouwillbehackedgoodluck

output

acbdefhijklmnogpqrstuvwxyz

题目给出一个字符串集，并假设是字典序的，要求26个字母的字典序（自定义的字典序），每两个字符串，可以得出一对字符的优先级，然后得出了所有字符的先后顺序，就转化成了拓扑排序的问题了，使用了优先队列，这样可以尽可能的小的在前面。其次，如果有ab a，这样的字符串，可以直接认为是不可能存在的！

#define N 105#define MOD 1000000000000000007struct node{    int x;    node(int xx){        x = xx;    }    bool operator < (const node a) const{        return x>a.x;    }};int n,in[30],ans[30],ansNum;char str[N][N];bool land[30][30];priority_queue<node> myqueue;bool getland(int x,int y){    int len = min(strlen(str[x]),strlen(str[y]));    FI(len){        if(str[x][i] != str[y][i]){            land[str[x][i] - 'a'][str[y][i]-'a'] = true;            return true;        }    }    if(strlen(str[x])>strlen(str[y]))    return false;    return true;}int main(){    while(S(n)!=EOF)    {        FI(n){            SS(str[i]);        }        memset(land,false,sizeof(land));        bool flag = true;        for(int i=0;i<n && flag;i++){            for(int j = i+1;j<n && flag;j++){                flag = getland(i,j);            }        }        if(!flag){            printf("Impossible\n");            continue;        }        memset(in,0,sizeof(in));        FI(26){            FJ(26){                if(land[i][j]){                    in[j]++;                }            }        }        while(!myqueue.empty())            myqueue.pop();        for(int i=0;i<26;i++){            if(in[i] == 0){               myqueue.push(node(i));            }        }        ansNum = 0;        while(!myqueue.empty()){            node top = myqueue.top();            myqueue.pop();            ans[ansNum++] = top.x;            FI(26){                if(land[top.x][i]){                    in[i]--;                    if(in[i] == 0)                    myqueue.push(node(i));                }            }        }        if(ansNum == 26){            FI(ansNum)                printf("%c",ans[i]+'a');            printf("\n");        }        else            printf("Impossible\n");    }    return 0;}