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  • 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 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
    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

    两两比较字符串。如果前一个串是当前串的前缀则无需重排字母表;如果当前串是前一个串的前缀则发生错误,Impossible;否则就找到最左不同字符,并形成两结点,令前一个字符结点指向当前字符结点,最后再做一次拓扑排序。

      1 /*
      2 拓扑排序,跑一个队列
      3 */
      4 #include <iostream>
      5 #include <cstdio>
      6 #include <cstring>
      7 #include <algorithm>
      8 #include <cmath>
      9 #include <string>
     10 #include <vector>
     11 #include <set>
     12 #include <map>
     13 #include <queue>
     14 #include <stack>
     15 using namespace std;
     16 const int INF = 0x7fffffff;
     17 const double EXP = 1e-8;
     18 const int MS = 105;
     19 const int cnt = 26;
     20 int n;
     21 char str[MS][MS];
     22 bool link[cnt][cnt];
     23 int in[cnt];   //入度
     24 char ans[cnt];
     25 bool have;
     26 void input()
     27 {
     28     cin >> n;
     29     for (int i = 0; i < n; i++)
     30         cin >> str[i];
     31 
     32     memset(in, 0, sizeof(in));
     33     memset(link, false, sizeof(false));
     34     have = true;
     35     for (int i = 0; i < n - 1&&have; i++)
     36     {
     37         int len1 = strlen(str[i]);
     38         int len2 = strlen(str[i + 1]);
     39         int ok = true;
     40         for (int j = 0; j < len1&&j < len2&&ok; j++)
     41         {
     42             if (str[i][j] != str[i + 1][j])
     43             {
     44                 ok = false;
     45                 if (!link[str[i][j] - 'a'][str[i + 1][j] - 'a'])
     46                 {
     47                     link[str[i][j] - 'a'][str[i + 1][j] - 'a'] = true;
     48                     in[str[i + 1][j] - 'a']++;
     49                 }
     50             }
     51         }
     52         if (ok&&len1 > len2)
     53         {
     54             have = false;
     55         }
     56     }
     57 }
     58 
     59 void solve()
     60 {
     61     if (!have)
     62     {
     63         cout << "Impossible" << endl;
     64         return;
     65     }
     66     queue<int > que;
     67     int num = 0;
     68     for (int i = 0; i < cnt; i++)
     69         if (in[i] == 0)
     70         {
     71             que.push(i);
     72             ans[num++] = 'a' + i;
     73         }
     74     while (!que.empty())
     75     {
     76         int s = que.front();
     77         que.pop();
     78         for (int i = 0; i < cnt; i++)
     79         {
     80             if (link[s][i])
     81             {
     82                 in[i]--;
     83                 if (in[i] == 0)
     84                 {
     85                     ans[num++] = 'a' + i;
     86                     que.push(i);
     87                 }
     88             }
     89         }
     90     }
     91     if (num < cnt)
     92         cout << "Impossible" << endl;
     93     else
     94     {
     95         ans[num] = '';
     96         cout << ans << endl;
     97     }
     98 }
     99 
    100 int main()
    101 {
    102     input();
    103     solve();
    104     return 0;
    105 }
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  • 原文地址:https://www.cnblogs.com/767355675hutaishi/p/4270434.html
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