思路:對相鄰兩行,當s1[i]≠s2[i]時,新的字母表中s1[i]必須出現在s2[i]的前面,顯然是拓撲排序。所以枚舉相鄰的兩行,構造有向邊統計入度。
拓撲排序:
每次從圖中拿走一個入度爲0的節點,並刪除所有從該點出發的有向邊。若最終輸出的節點個數小於總個數,證明存在環。
代碼如下:
#include <cstdio>
#include <cstring>
using namespace std;
#define N 105
char s[N][N], ans[27];
int head[26], degree[26];
struct Edge{
int to, next;
}e[700];
int main(){
int n;
scanf("%d", &n);
for(int i = 0; i < n; ++i)
getchar(), scanf("%s", s[i]);
int cnt = 0;
bool failed = false;
for(int i = 1; i < n && !failed; ++i){
int len1 = strlen(s[i - 1]), len2 = strlen(s[i]);
int j = 0;
for(; j < len1 && j < len2; ++j){
int u = s[i - 1][j] - 'a', v = s[i][j] - 'a';
if(u == v)
continue;
++cnt;
e[cnt].to = v, e[cnt].next = head[u], head[u] = cnt, ++degree[v];
break;
}
if(j == len2 && j < len1)
failed = true;
}
if(failed)
printf("Impossible\n");
else{
memset(ans, 0, sizeof(ans));
bool single;
int idx = 0;
do{
single = false;
for(int i = 0; i < 26 && !single; ++i){
if(degree[i] == 0){
single = true;
ans[idx ++] = i + 'a';
--degree[i];
for(int j = head[i]; j != 0; j = e[j].next)
--degree[e[j].to];
}
}
}while(single);
if(idx < 26)
printf("Impossible\n");
else
printf("%s\n", ans);
}
return 0;
}