（CodeForces 510C） Fox And Names 拓扑排序

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 ti 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).

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

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <math.h>
#include <algorithm>
#include <queue>
using namespace std;

#define met(a,b) memset(a,b,sizeof(a))
#define ll long long
#define N 123
struct node
{
    int v,next;
}s[N];
int a[N],num[N],vis[N],k;
int len[N],in[N];
void add(int u,int v)
{
    s[k].v=u;
    s[k].next=a[v];
    a[v]=k++;
    in[u]++;
}
char str[N][N];

int main()
{
          int n;
       scanf("%d",&n);met(a,-1);k=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",str[i]);
            len[i]=strlen(str[i]);
        }
            met(in,0);int j;
        for(int i=1;i<n;i++)
        {
            for(j=0;j<min(len[i],len[i-1]);j++)
            {
                if(str[i][j]!=str[i-1][j])///两个字符串的相同位置比较
                {
                    int x=str[i][j]-‘a‘;
                    int y=str[i-1][j]-‘a‘;
                    add(x,y);
                    break;
                }
            }
            if(len[i]<len[i-1] && j==len[i])///前一个字符串比后一个长，输出不可能
            {
                printf("Impossible\n");
                return 0;

            }
        }
        met(vis,0);
        for(int i=1;i<=26;i++)
        {
            int jj=26;
            int j;
            for(int j=0;j<26;j++)///一个一个字符查找符合的
            {
                if(!vis[j] && in[j]==0)
                {
                    vis[j]=1;
                    num[i]=j;
                    for(int k=a[j];k!=-1;k=s[k].next)
                        in[s[k].v]--;

                jj=j;
                break;
                }
            }
            if(jj==26)///中间有断开的，有一个字符没有比较
            {
                printf("Impossible\n");
                return 0;
            }
        }
        for(int i=1;i<=26;i++)
            printf("%c",num[i]+‘a‘);
        puts("");

    return 0;
}