Codeforces 290C

        题意：给出一组没有按字母表排序的英文名字，要求找出一种新的字母表顺序，使这组名字符合该字母表序。

        解法：拓扑排序的应用，注意判断 xy 在 x 之前的情况。

// Codeforces 290C
#include<iostream>
#include<cstdio>
#include<cstring>
#include<memory.h>
using namespace std;

class alphaGraph{
private:
	struct edgeNode{
		char end;
		edgeNode* next;

		edgeNode(char e, edgeNode* n = NULL) : end(e), next(n) {}
		edgeNode() : next(NULL) {}
	};

	struct verNode{
		//char ch;          // Need it?
		edgeNode* head;

		verNode() : head(NULL) {}
	};

	verNode table[26];

public:
	char alphabet[30];

	alphaGraph(){}
	~alphaGraph(){
		int i;
		for (i = 0; i <= 25; i++){
			edgeNode* tmp = table[i].head;
			while (tmp){
				table[i].head = table[i].head->next;
				delete tmp;
				tmp = table[i].head;
			}
		}
	}

	void insert(char s, char e){
		table[s - 'a'].head = new edgeNode(e, table[s - 'a'].head);
	}

	bool exist(char s, char e){
		edgeNode* tmp = table[s - 'a'].head;
		while (tmp){
			if (tmp->end == e) return true;
			tmp = tmp->next;
		}
		return false;
	}

	bool checkLoop(){             // Apply the topological sort to find if loop exits
		int i, cnt = 0, queue[100], front = 0, rear = 0, deg[26];
		memset(deg, 0, sizeof(deg));
		for (i = 0; i <= 25; i++) {      // Calculate the degrees
			edgeNode* tmp = table[i].head;
			while (tmp){
				deg[tmp->end - 'a']++;
				tmp = tmp->next;
			}
		}

		for (i = 0; i <= 25; i++)
		if (deg[i] == 0) queue[rear++] = i;

		// Dequeue the node with degree of 0, and calculate the degrees again in a loop
		while (rear != front){          
			int cur = queue[front++];
			alphabet[cnt++] = cur + 'a';

			edgeNode* tmp = table[cur].head;
			while (tmp){
				deg[tmp->end - 'a']--;
				if (deg[tmp->end - 'a'] == 0) queue[rear++] = tmp->end - 'a';
				tmp = tmp->next;
			}
		}

		if (cnt == 26) {
			alphabet[26] = '\0';
			return false;
		}
		else return true;
	}
};

char name[110][110];

// Return true if there is a probable order
bool check(int st, int ed, int level, alphaGraph& gp){
	int i;
	for (i = st; i <= ed; i++){
		// 找到下一个不相同的字符
		while (i < ed && name[i + 1][level] == name[i][level]) i++;
		if (i == ed) continue;
	    // x < xy
		if (!name[i][level] && name[i + 1][level]) continue;
		// xy < x
		if (name[i][level] && !name[i + 1][level]) return false;
		// 防止出现重边
		if (!gp.exist(name[i][level], name[i + 1][level]))
			gp.insert(name[i][level], name[i + 1][level]);
	}

	if (gp.checkLoop()) return false;

	// 检查下一层
	int next_ed;
	for (i = st; i <= ed; i++){
		if (name[i][level] == '\0') continue;
		if (i < ed && name[i][level] == name[i + 1][level]){
			next_ed = i;
			while (next_ed < ed && name[next_ed][level] == name[next_ed + 1][level])
				next_ed++;
			if (!check(i, next_ed, level + 1, gp)) return false;
			i = next_ed;
		}
	}

	return true;
}

int main(){
	int n, i;
	cin >> n;
	for (i = 0; i <= n - 1; i++) cin >> name[i];

	alphaGraph gp;
	if (check(0, n - 1, 0, gp)) cout << gp.alphabet << endl;
	else cout << "Impossible" << endl;

	return 0;
}