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).
Example 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
改變26個字母的順序使得給出的字符串是按字典序排列下來的,i+1字符串的j位置的字母a如果不等于i字符串j位置的字母b,那么字母a的順序就得在b的后面,轉化成圖的話就是b有一條有向邊連到a(a的拓撲順序在b后),如果最終處理完所有字符串后建立的圖出現了環(拓撲排序后有的點不能加入答案隊列)或者字符串x比字符串y長且x的前leny位與y相同,即為impossible,否則輸出拓撲排序后的結果。
#include<algorithm>#include<string>#include<iostream>#include<stack>#include<vector>#include<queue>using namespace std;const int N = 200;string now,pre;int deg[26],ans[26];vector<int> g[26];queue<int> que;int main(){ int n,tot = 0; cin >> n; for(int i=0; i<n; i++) { cin >> now; int len = min(now.length(),pre.length()); bool ok = false; for(int j=0; j<len; j++) { if(now[j] != pre[j]) { ok = true; g[pre[j]-'a'].push_back(now[j]-'a'); deg[now[j]-'a']++; break; } } if(!ok&&now.length() < pre.length()) { cout << "Impossible" << endl; return 0; } pre=now; } for(int i=0; i<26; i++) if(!deg[i]) que.push(i); while(!que.empty()) { int x = que.front(); ans[tot++] = x; que.pop(); for(int i=0; i<g[x].size(); i++) { if(deg[g[x][i]] == 1) que.push(g[x][i]); if(deg[g[x][i]] != 0) deg[g[x][i]]--; } } if(tot != 26) cout << "Impossible" << endl; else { for(int i=0; i<26; i++) { cout <<char('a'+ ans[i]); } cout << endl; } return 0;}新聞熱點
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