C. Fox And Names
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 inlexicographical 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 tiaccording
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 test(s)
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
题意：给n个字符串，它们按照某个字典序从小到大排列，问这个字典序是否存在，存在就输出任意一个满足条件的字典序，否则输出“Impossible”。
裸的topsort，结果在终判时挂了，就因为没有特判，杯具。。。。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#define maxn 111
using namespace std;
int indegree[maxn],path[maxn];
int n,m;
char w[105][105];
queue<int>Q;
vector<int>edges[10000];
void topsort()
{
memset(path,0,sizeof(path));
while (!Q.empty())
Q.pop();
for (int i=0;i<26;i++)
if (!indegree[i])
Q.push(i);
int num=0;
while (!Q.empty())
{
int now=Q.front();
Q.pop();
path[num++]=now;
for (int i=0;i<edges[now].size();i++)
{
--indegree[ edges[now][i]];
if (!indegree[ edges[now][i] ])
{
Q.push(edges[now][i]);
}
}
}
if (num<26)
{
printf("Impossible\n");
return ;
}
for (int i=0;i<num;i++)
printf("%c",path[i]+'a');
printf("\n");
}
int main()
{
while (~scanf("%d",&n))
{
memset(indegree,0,sizeof(indegree));
for (int i=1;i<=n;i++)
edges[i].clear();
int u,v;
for (int i=0;i<n;i++)
scanf("%s",w[i]);
int flag=0;
for (int i=1;i<n;i++)
{
int a=i-1,b=i;
int la=strlen(w[a]);
int lb=strlen(w[b]);
int aa=0,bb=0;
while (aa<la&&bb<lb&&w[a][aa]==w[b][bb])
{
aa++;
bb++;
}
if (bb==lb&&aa<la) {flag=1; break;} //掉了这一句，结果挂了。。。
if (aa==la&&bb<=lb) continue;
indegree[w[b][bb]-'a']++;
edges[w[a][aa]-'a'].push_back(w[b][bb]-'a');
}
if (flag)
{
printf("Impossible\n");
continue;
}
topsort();
}
return 0;
}