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 siand 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).
Sample test(s)
input
3rivestshamiradleman
output
bcdefghijklmnopqrsatuvwxyz
input
10touristpetrwjmzbmryeputonsvepifanovscottwuoooooooooooooooosubscriberrowdarktankengineer
output
Impossible
input
10petregorendagorionfeferivanilovetanyaromanovakostkadmitriyhmaratsnowbearbredorjaguarturnikcgyforever
output
aghjlnopefikdmbcqrstuvwxyz
input
7carcarecarefulcarefullybecarefuldontforgetsomethingotherwiseyouwillbehackedgoodluck
output
acbdefhijklmnogpqrstuvwxyz
拓扑排序第一题，注意特判一下下面的串是上面的串的前缀这种情况，直接输出Impossible了。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 bool G[30][30];
5 int IN[30];
6 int ANS[30];
7
8 void toposort(void);
9 int main(void)
10 {
11 int n;
12 char name[105][105];
13 bool flag;
14
15 cin >> n;
16 for(int i = 0;i < n;i ++)
17 cin >> name[i];
18 for(int i = 0;i < n - 1;i ++)
19 {
20 for(int j = 0;name[i][j] && name[i + 1][j];j ++)
21 {
22 flag = false;
23 if(name[i][j] != name[i + 1][j])
24 {
25 if(!G[name[i][j] - ‘a‘][name[i + 1][j] - ‘a‘])
26 {
27 G[name[i][j] - ‘a‘][name[i + 1][j] - ‘a‘] = true;
28 IN[name[i + 1][j] - ‘a‘] ++;
29 }
30 flag = true;
31 break;
32 }
33 }
34 if(!flag && strlen(name[i]) > strlen(name[i + 1]))
35 {
36 puts("Impossible");
37 return 0;
38 }
39 }
40 toposort();
41
42 return 0;
43 }
44
45 void toposort(void)
46 {
47 queue<int> que;
48 int sum = 1;
49 int k = 0;
50 int front;
51
52 for(int i = 0;i < 26;i ++)
53 if(!IN[i])
54 {
55 ANS[k ++] = i;
56 sum ++;
57 que.push(i);
58 }
59
60 while(!que.empty())
61 {
62 front = que.front();
63 que.pop();
64 for(int i = 0;i < 26;i ++)
65 if(G[front][i])
66 {
67 IN[i] --;
68 if(!IN[i])
69 {
70 ANS[k ++] = i;
71 que.push(i);
72 sum ++;
73 }
74 }
75 }
76 if(sum < 26)
77 puts("Impossible");
78 else
79 {
80 for(int i = 0;i < 26;i ++)
81 printf("%c",ANS[i] + ‘a‘);
82 puts("");
83 }
84
85 return ;
86 }