Description

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 in lexicographical 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 si and 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 Input

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
标准的拓扑排序。
 #include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
char s[][];
queue<int> q;
int ans[],in[];
int g[][];
int cnt;
void toposort(){
queue<int> q;
for(int i = ; i<=; i++)
if(!in[i]) q.push(i);
cnt = ;
while(!q.empty()){
int tmp = q.front();
q.pop();
ans[++cnt] = tmp;
for(int j = ; j<=; j++){
if(g[tmp][j]){
in[j]--;
if(!in[j]) q.push(j);
}
}
}
}
void input(){
int n;
scanf("%d",&n);
for(int i = ; i<=n; i++) scanf("%s",s[i]);
bool flag1 = true;
for(int i = ; i<n; i++){
int lena = strlen(s[i]);
int lenb = strlen(s[i+]);
char a,b;
int tem1,tem2;
int flag = ;
for(int j = ; j<min(lena,lenb); j++){
if(s[i][j]!=s[i+][j]){
a = s[i][j];
b = s[i+][j];
tem1 = a - 'a' + ;
tem2 = b - 'a' + ;
if(!g[tem1][tem2]){ // 这没考虑到已经有路径的条件,一直WA
g[tem1][tem2] = ;
in[tem2]++;
}
flag = ;
break;
}
}
if(!flag&&lena>lenb){
flag1 = false;
break;
}
}
if(!flag1) printf("Impossible\n");
else{
toposort();
if(cnt<) {
printf("Impossible\n");
return;
}
for(int i = ; i<=; i++){
printf("%c",ans[i]-+'a');
}
printf("\n");
}
}
int main()
{
input();
return ;
}/*
26
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
y
z
*/

卷珠帘

Fox And Names的更多相关文章

  1. Codeforces Round #290 (Div. 2) C. Fox And Names dfs

    C. Fox And Names 题目连接: http://codeforces.com/contest/510/problem/C Description Fox Ciel is going to ...

  2. C. Fox And Names

    C. Fox And Names time limit per test 2 seconds memory limit per test 256 megabytes input standard in ...

  3. CF Fox And Names (拓扑排序)

    Fox And Names time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  4. codeforce 510C Fox And Names(拓扑排序)

    Fox And Names time limit per test 2 seconds memory limit per test 256 megabytes input standard input ...

  5. (CodeForces 510C) Fox And Names 拓扑排序

    题目链接:http://codeforces.com/problemset/problem/510/C Fox Ciel is going to publish a paper on FOCS (Fo ...

  6. [CF #290-C] Fox And Names (拓扑排序)

    题目链接:http://codeforces.com/contest/510/problem/C 题目大意:构造一个字母表,使得按照你的字母表能够满足输入的是按照字典序排下来. 递归建图:竖着切下来, ...

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

    题目传送门 /* 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 拓扑排序 详细解释:http://www.2cto.com/kf/201502/374966.html */ #i ...

  8. codeforces 510 C Fox And Names【拓扑排序】

    题意:给出n串名字,表示字典序从小到大,求符合这样的字符串排列的字典序 先挨个地遍历字符串,遇到不相同的时候,加边,记录相应的入度 然后就是bfs的过程,如果某一点没有被访问过,且入度为0,则把它加入 ...

  9. CodeForces 510C Fox And Names (拓扑排序)

    <题目链接> 题目大意: 给你一些只由小写字母组成的字符串,现在按一定顺序给出这些字符串,问你怎样从重排字典序,使得这些字符串按字典序排序后的顺序如题目所给的顺序相同. 解题分析:本题想到 ...

随机推荐

  1. Git的Bug分支----临时保存现场git stash

    软件开发中,bug就像家常便饭一样,有了bug就需要修复,在Git中,由于分支是如此的强大,所以每个bug通过一个新的分支来修复,在修复后,合并分支,然后将临时分支删除. 当你接到一个修复代号为119 ...

  2. Git 怎么创建本地库,向本地库提交文件

    创建版本库是非常简单的,首先选择一个自己想放的位置,创建一个空目录: (用windows的git bash,这个工具的操作命令和linux下类似) $ mkdir gitRespository     ...

  3. DataBinding

    <?xml version="1.0" encoding="utf-8"?> <layout xmlns:android="http ...

  4. bfs UESTC 381 Knight and Rook

    http://acm.uestc.edu.cn/#/problem/show/381 题目大意:给你两个棋子:车.马,再给你一个n*m的网格,从s出发到t,你可以选择车或者选择马开始走,图中有一些障碍 ...

  5. c_select 调用参数说明

    c_select 调用 1. select系统调用select系统调用是用来让我们的程序监视多个文件描述符的状态变化的.程序会停在select这里等待,直到被监视的文件描述符有某一个或多个发生了状态改 ...

  6. isr

    Kafaka动态维护了一个同步状态的副本的集合(a set of in-sync replicas),简称ISR

  7. 用sql server的sql语句算一个empty GUID

    在C#中得到一个empty GUID的方法是: Guid id= Guid.Empty; 那么在SQL Server Management Studio中怎样得到一个empty GUID呢? 方法有两 ...

  8. git repository 的使用

    1. You should only be pushing to a bare repository. A bare repository is a repository that has no ch ...

  9. 转 由一次磁盘告警引发的血案:du 和 ls 的区别

    如果你完全不明白或者完全明白图片含义, 那么你不用继续往下看了. 否则, 这篇文章也许正是你需要的. 背景 确切地说,不是收到的自动告警短信或者邮件告诉我某机器上的磁盘满了,而是某同学人肉发现该机器写 ...

  10. 转 awk 使用方法

    Shell编程-awk 简介 awk 是一种对立的编程语言,集成于所有UNIX/Linux中,这个名字是它创建者的名字首字母组成的 Alfred Aho,Peter Weinberger, and B ...