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

 

题目传送门

 1 /*  2  给出n个字符串，求是否有一个“字典序”使得n个字符串是从小到大排序  3  拓扑排序  4  详细解释：http://www.2cto.com/kf/201502/374966.html  5 */  6 #include <cstdio>  7 #include <iostream>  8 #include <cstring>  9 #include <string>  10 #include <map>  11 #include <algorithm>  12 #include <vector>  13 #include <set>  14 #include <queue>  15 #include <cmath>  16 using namespace std;  17  18 const int MAXN = 1e6 + 10;  19 const int INF = 0x3f3f3f3f;  20 char s[110][110];  21 int in[110];  22 bool lin[27][27];  23 char ans[27];  24  25 bool TopoSort(void)  26 {  27 queue<int> q;  28 int cnt = 0;  29 for (int i=1; i<=26; ++i)  30  {  31 if (in[i] == 0)  32  {  33  q.push (i);  34 ans[cnt++] = 'a' + i - 1;  35  }  36  }  37 while (!q.empty ())  38  {  39 int now = q.front (); q.pop ();  40 for (int i=1; i<=26; ++i)  41  {  42 if (lin[now][i])  43  {  44 in[i]--;  45 if (in[i] == 0)  46  {  47  q.push (i);  48 ans[cnt++] = 'a' + i - 1;  49  }  50  }  51  }  52  }  53 ans[26] = '\0';  54 if (cnt < 26) return false;  55 else return true;  56 }  57  58 int main(void)  59 {  60 //freopen ("C.in", "r", stdin);  61  62 int n;  63  64 while (~scanf ("%d", &n))  65  {  66 memset (lin, false, sizeof (lin));  67 memset (in, 0, sizeof (in));  68 for (int i=1; i<=n; ++i)  69  {  70 scanf ("%s", &s[i]);  71  }  72  73 bool flag = true;  74 for (int i=1; i<=n-1 && flag; ++i)  75  {  76 int len1 = strlen (s[i]);  77 int len2 = strlen (s[i+1]);  78 bool ok = false;  79 for (int j=0; j<=len1-1 && j<=len2-1 && !ok; ++j)  80  {  81 if (s[i][j] != s[i+1][j])  82  {  83 ok = true;  84 if (!lin[s[i][j]-'a'+1][s[i+1][j]-'a'+1])  85  {  86 in[s[i+1][j]-'a'+1]++;  87 lin[s[i][j]-'a'+1][s[i+1][j]-'a'+1] = true;  88  }  89  }  90  }  91 if (!ok && len1 > len2) flag = false;  92  }  93  94 if (!flag) puts ("Impossible");  95 else  96  {  97 flag = TopoSort ();  98 if (!flag) puts ("Impossible");  99 else printf ("%s\n", ans); 100  } 101  } 102 103 return 0; 104 }