【题目链接】:http://codeforces.com/contest/510/problem/C

【题意】



给你n个字符串;

问你要怎么修改字典序;

(即原本是a,b,c..z现在你可以修改每个字母在字典序中的位置了);

才能使得这n个字符串是字典序升序的;

【题解】



对于第i个字符串;

让他和第i+1..n个字符串比较->j;

找到第一个不同的位置pos;

然后

必然是要

s[i][pos]< s[j][pos]的；

拓扑排序!

即s[i][pos]建一条边指向s[j][pos]

然后做拓扑排序就好;

如果某个串是另外一个串的子串;

则如果s[i]是s[j]的子串,那么可行,否则无解;



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define ps push_back

#define fi first

#define se second

#define rei(x) cin >> x

#define pri(x) cout << x

#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int MAXC = 26+10;

const int N = 100+10;

int a[MAXC][MAXC],n,du[MAXC],cnt=0;

queue <int> dl;

string s[N];

vector <int> ans;

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    ios::sync_with_stdio(false);

    rei(n);

    rep1(i,1,n)

        rei(s[i]);

    rep1(i,1,n)

    {

        rep1(j,i+1,n)

        {

            int ma = min(int(s[i].size()),int(s[j].size()));

            int pos = -1;

            rep1(k,0,ma-1)

                if (s[i][k]!=s[j][k])

                {

                    pos = k;

                    break;

                }

            if (pos==-1)

            {

                if (int(s[i].size())>int(s[j].size()))

                    return pri("Impossible"<<endl),0;

                else

                    continue;

            }

            int x = s[i][pos]-'a'+1,y = s[j][pos]-'a'+1;

            if (a[x][y]) continue;

            du[y]++;

            a[x][y]=1;

        }

    }

    rep1(i,1,26)

        if (du[i]==0)

            dl.push(i),cnt++,ans.ps(i);

    while (!dl.empty())

    {

        int x = dl.front();

        dl.pop();

        rep1(j,1,26)

            if (a[x][j])

            {

                a[x][j] = 0;

                du[j]--;

                if (du[j]==0)

                {

                    dl.push(j);

                    ans.ps(j);

                    cnt++;

                }

            }

    }

    if (cnt==26)

        rep1(i,0,25)

            putchar(ans[i]+'a'-1);

    else

        return pri("Impossible"<<endl),0;

    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);

    return 0;

}