codeforce 510C Fox And Names（拓扑排序）

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 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

Copy

3
rivest
shamir
adleman

output

Copy

bcdefghijklmnopqrsatuvwxyz

input

Copy

10
tourist
petr
wjmzbmr
yeputons
vepifanov
scottwu
oooooooooooooooo
subscriber
rowdark
tankengineer

output

Copy

Impossible

input

Copy

10
petr
egor
endagorion
feferivan
ilovetanyaromanova
kostka
dmitriyh
maratsnowbear
bredorjaguarturnik
cgyforever

output

Copy

aghjlnopefikdmbcqrstuvwxyz

input

Copy

7
car
care
careful
carefully
becarefuldontforgetsomething
otherwiseyouwillbehacked
goodluck

output

Copy

acbdefhijklmnogpqrstuvwxyz


大坑点：
2
aa
a
这应该是impossible

题意：有n个按照某一字母表排好序的字符串，问是否存在这样的字母表。或者说是否存在这样一张字母表使字符串排序后是这个样子。

 

思路：可以根据字符串的顺序得出一些图的指向，比如abc和abd，可以得到c->d，这样就可以得到若干边，根据这个进行拓扑排序即可。注意有个陷阱是如果存在xy>x，那就无解。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
char a[105][105];
int in[30];
int out[30];
bool book[30];
vector<int>v[35];
queue<int>q;
int main()
{
    int n;scanf("%d",&n);
    memset(book,0,sizeof(book));
    memset(in,0,sizeof(0));
    memset(out,0,sizeof(0));
    while(!q.empty()) q.pop();
    for(int i=1;i<=n;i++)scanf("%s",&a[i]);
    for(int i=1;i<=n;i++)
    {

        for(int j=i+1;j<=n;j++)
        {
            int l=min(strlen(a[i]),strlen(a[j]));
            for(int k=0;k<l;k++)
            {
                if(a[i][k]!=a[j][k])
                {
                    in[a[j][k]-'a'+1]++;//保存入度
                    out[a[i][k]-'a'+1]++;//其实不用考虑出度的
                    v[a[i][k]-'a'+1].push_back(a[j][k]-'a'+1);
                    break;
                }
                if(k==l-1&&strlen(a[i])>strlen(a[j]))
                {
                    printf("Impossible");
                    return 0;
                }
            }
        }
    }
    //cout<<in['e'-'a'+1]<<endl;
   // cout<<in['p'-'a'+1]<<endl;
    int k=0;
    while(1)
    {
        int x=-1;
        for(int i=1;i<=26;i++)
        {
            if(in[i]==0&&!book[i])//找到没有入度的点
            {
                x=i;
                break;
            }
        }
        if(x==-1&&k<26)
        {
            //cout<<k<<endl;
            cout<<"Impossible";
            break;
        }
        if(x==-1) break;
        k++;
        book[x]=1;
        q.push(x);
        for(int i=0;i<v[x].size();i++)//将挑选出来的改点的边都去掉
        {
            in[v[x][i]]--;
        }
    }
    if(k==26)
    {
        while(!q.empty())
        {
            char t=q.front()+'a'-1;
            cout<<t;
            q.pop();
        }
    }
    return 0;
}