C. Fox And Names Codeforces Round #290 (Div. 二)

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

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

题意：给n个字符串，它们按照某个字典序从小到大排列，问这个字典序是否存在，存在就输出任意一个满足条件的字典序，否则输出“Impossible”。

裸的topsort，结果在终判时挂了，就因为没有特判，杯具。。。。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#define maxn 111
using namespace std;

int indegree[maxn],path[maxn];
int n,m;
char w[105][105];
queue<int>Q;
vector<int>edges[10000];

void topsort()
{
    memset(path,0,sizeof(path));
    while (!Q.empty())
        Q.pop();
    for (int i=0;i<26;i++)
        if (!indegree[i])
            Q.push(i);
    int num=0;
    while (!Q.empty())
    {
        int now=Q.front();
        Q.pop();
        path[num++]=now;
        for (int i=0;i<edges[now].size();i++)
        {
            --indegree[ edges[now][i]];
            if (!indegree[ edges[now][i] ])
            {
                Q.push(edges[now][i]);
            }
        }
    }
    if (num<26)
    {
        printf("Impossible\n");
        return ;
    }
    for (int i=0;i<num;i++)
        printf("%c",path[i]+'a');
    printf("\n");
}

int main()
{
    while (~scanf("%d",&n))
    {
        memset(indegree,0,sizeof(indegree));
        for (int i=1;i<=n;i++)
            edges[i].clear();
        int u,v;
        for (int i=0;i<n;i++)
            scanf("%s",w[i]);
        int flag=0;
        for (int i=1;i<n;i++)
        {
            int a=i-1,b=i;
            int la=strlen(w[a]);
            int lb=strlen(w[b]);
            int aa=0,bb=0;
            while (aa<la&&bb<lb&&w[a][aa]==w[b][bb])
            {
                aa++;
                bb++;
            }
            if (bb==lb&&aa<la) {flag=1; break;} //掉了这一句，结果挂了。。。
            if (aa==la&&bb<=lb) continue;
            indegree[w[b][bb]-'a']++;
            edges[w[a][aa]-'a'].push_back(w[b][bb]-'a');
        }
        if (flag)
        {
            printf("Impossible\n");
            continue;
        }
        topsort();
    }
    return 0;
}

C. Fox And Names Codeforces Round #290 (Div. 二)

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