Codeforces Round #290(Div.2) C. Fox And Names(토폴로지 정렬)

제목 링크:제목 링크
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

제목: n개의 문자열을 드리겠습니다. 이 n개의 문자열의 크기를 승차순으로 할 수 있도록 문자표의 크기 관계를 어떻게 사용자 정의해야 하는지 물어보십시오.
예를 들면 다음과 같습니다.
abc
aad
만약 그것들 두 개가 승차순이라면 사용자 정의 b는 틀림없이 a보다 크다.
그리고 조건에 맞는 알파벳을 출력합니다.
문제 풀이:
사실 문제는 비교적 간단하기 때문에 토폴로지 정렬을 이용하여 이 조작을 실현할 수 있다.
그러나 문자열의 비교가 어떠한지 주의해야 한다.
서로 다른 것을 만나면 그 비교값을 되돌려주고, 뒤의 것은 비교할 필요가 없다
토폴로지 정렬의 서열을 출력할 수 없으면 실현할 수 없습니다.
코드:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<vector>
#include<bitset>
#include<set>
#include<queue>
#include<stack>
#include<map>
#include<cstdlib>
#include<cmath>
#define PI 2*asin(1.0)
#define LL __int64
const int  MOD = 1e9 + 7;
const int N = 1e2 + 15;
const int INF = (1 << 30) - 1;
const int letter = 130;
using namespace std;
int n;
char str[N][N];
int mp[N][N];
bool dis[N][N];
int in[N], out[N];
bool vis[N];
queue<int>q;
vector<char>vs;
int value(char aim)
{
    if(aim == ' ') return 0;
    else return aim - 'a' + 1;
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        memset(vis, 0, sizeof(vis));
        memset(mp, 0, sizeof(mp));
        memset(in, 0, sizeof(in));
        memset(out, 0, sizeof(out));
        memset(dis, 0, sizeof(dis));
        getchar();
        int max1 = -1, len;
        for(int i = 0; i < n; i++)
        {
            gets(str[i]);
            len = strlen(str[i]);
            if(len > max1) max1 = len;
        }
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < max1; j++)
            {
                if(!(str[i][j] >= 'a' && str[i][j] <= 'z')) str[i][j] = ' ';
            }
            str[i][max1] = '\0';
        }
        int a, b;
        int t = 0;
        for(int j = 0; j < max1; j++)
        {
            for(int i = 0; i < n - 1; i++)
            {
                if(!dis[i][i + 1]) ///last
                {
                    a = value(str[i][j]);
                    b = value(str[i + 1][j]);
                    if(a != b)
                    {
                        dis[i][i + 1] = 1;
                        t++;
                       // in[b]++;
                        mp[a][b] = 1;
                     ///   printf("%d %d
", a, b);
                    }
                    if(t == n - 1) break; ///bijiao n-1ci
                }
            }
            if(t == n - 1) break;
        }
        for(int i = 0;i <= 26;i++)
            for(int j = 0;j <= 26;j++)
        {
            if(mp[i][j]) in[j]++;
        }
        while(!q.empty()) q.pop();
        vs.clear();
        for(int i = 00; i <= 26; i++) if(in[i] == 0) q.push(i);
        if(in[0])
        {
            puts("Impossible");
            continue;
        }
        int flag = 1;
        while(!q.empty())
        {
            int x = q.front();
            q.pop();
            vs.push_back(x);
            for(int i = 0; i <= 26; i++)
            {
                if(mp[x][i])
                {
                    in[i]--;
                    if(in[i] == 0 && i != 0) q.push(i);
                    out[x]--;
                }
            }
        }

        for(int i = 1; i <= 26; i++)
        {
            if(in[i])
            {
                flag = 0;
                break;
            }
        }

        if(flag == 0) puts("Impossible");
        else
        {
            for(int i = 0; i < vs.size(); i++)
            {
                if(vs[i] != 0)
                {
                    printf("%c", vs[i] + 'a' - 1);
                    vis[vs[i]] = 1;
                }

            }
            for(int i = 1; i <= 26; i++)
            {
                if(!vis[i]) printf("%c", i + 'a' - 1);
            }
            printf("
");
        }
    }
    return 0;
}