CF C. Fox And Names(拓扑排序)
题意：给出n个字符串，求是否存在一种新的字母大小顺序，使这n个字符串满足这种顺序。
思路：注意每相邻两个串相比较时，从头到尾，只要比较到相同位置，字母不同，就结束了，不要继续向后比较了。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<map> 9 #include<iomanip> 10 #include<climits> 11 #include<string.h> 12 #include<numeric> 13 #include<cmath> 14 #include<stdlib.h> 15 #include<vector> 16 #include<stack> 17 #include<set> 18 #define FOR(x, b, e) for(int x=b;x<=(e);x++) 19 #define REP(x, n) for(int x=0;x<(n);x++) 20 #define INF 1e7 21 #define MAXN 100010 22 #define maxn 1000010 23 #define Mod 1000007 24 #define N 110 25 using namespace std; 26 typedef long long LL; 27 28 int n; 29 string s[N]; 30 int G[N][N]; 31 int ind[N]; 32 bool flag; 33 string ans; 34 35 bool gao() 36 { 37 queue<int> q; 38 for (int i = 0;i < 26;++ i) 39 if (ind[i] == 0) { 40 q.push(i); 41 ans += (char)(i + 'a'); 42 } 43 while (!q.empty()) { 44 int t = q.front(); 45 q.pop(); 46 for (int i = 0;i < 26;++ i) 47 if (G[t][i]) { 48 if (--ind[i] == 0) { 49 q.push(i); 50 ans += (char)(i + 'a'); 51 } 52 } 53 } 54 if (ans.length() == 26) return 1; 55 return 0; 56 } 57 58 int main() 59 { 60 cin >> n; 61 for (int i = 1;i <= n;++ i) 62 cin >> s[i]; 63 for (int i = 2;i <= n;++ i) { 64 int len1 = s[i - 1].length(); 65 int len2 = s[i].length(); 66 bool ok = false; 67 for (int j = 0;j < len1 && j < len2 && !ok;++ j) { 68 if (s[i-1][j] != s[i][j]) { 69 ok = true; 70 if (G[s[i-1][j] - 'a'][s[i][j] - 'a'] == 0) { 71 G[s[i-1][j] - 'a'][s[i][j] - 'a'] = 1; 72 ind[s[i][j] - 'a']++; 73 } 74 } 75 } 76 if (!ok && len1 > len2) { 77 puts("Impossible"); 78 return 0; 79 } 80 } 81 //cout << "ok" << endl; 82 int res = gao(); 83 if (res == 0) puts("Impossible"); 84 else cout << ans << endl; 85 return 0; 86 }
