Codeforces 510C Fox And Names 拓扑排序

传送门：510C

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 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

3rivestshamiradleman

output

bcdefghijklmnopqrsatuvwxyz

input

10touristpetrwjmzbmryeputonsvepifanovscottwuoooooooooooooooosubscriberrowdarktankengineer

output

Impossible

input

10petregorendagorionfeferivanilovetanyaromanovakostkadmitriyhmaratsnowbearbredorjaguarturnikcgyforever

output

aghjlnopefikdmbcqrstuvwxyz

input

7carcarecarefulcarefullybecarefuldontforgetsomethingotherwiseyouwillbehackedgoodluck

output

acbdefhijklmnogpqrstuvwxyz

解题方法：先把字典字母之间的关系找到。通过图连接起来。然后用拓扑排序

如果拓扑排序还没有学过。可以看http://www.cnblogs.com/newpanderking/archive/2012/10/18/2729552.html   这里面讲的很好

#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<cmath>#include<queue>#define INF 0x0f0f0f0fusing namespace std;char s[105][105],ans[30];int cou[30];bool link[30][30];queue<int> q;bool solve() //拓扑排序 {int i,j,k,l,font,cnt=0;for(i=0;i<26;i++){if(!cou[i]){q.push(i);ans[cnt++]='a'+i;}}while(!q.empty()){font=q.front();q.pop();for(i=0;i<26;i++){if(link[font][i]){cou[i]--;if(cou[i]==0){q.push(i);ans[cnt++]='a'+i;}}}}ans[26]='\0';if(cnt<26)return false;else return true;}int main(){int i,j,k,l,m,n,l1,l2,jud;k=1;memset(link,0,sizeof(link));scanf("%d%*c",&n);for(i=0;i<n;i++){scanf("%s",&s[i]);}for(i=0;i<n-1;i++){l1=strlen(s[i]);l2=strlen(s[i+1]);jud=0;for(j=0;j<l1&&j<l2&&!jud;j++){if(s[i][j]!=s[i+1][j]){jud=1;if(!link[s[i][j]-'a'][s[i+1][j]-'a']){link[s[i][j]-'a'][s[i+1][j]-'a']=true;cou[s[i+1][j]-'a']++;}}}if(!jud&&l1>l2) k=0;//某两个字符串前缀完全相同，但是第一个字符串长度较大，则两字符串必定不满足字典序}if(!k){printf("Impossible\n");}else{k=solve();if(!k)printf("Impossible\n");else printf("%s\n",ans);}return 0;}