Codeforces Round #445 div.2 D. Restoration of string 乱搞

D. Restoration of string

题意：给你n个字符串，让你构造一个终串，使得这n个字符串都是终串的最小频繁子串，如果不存在输出NO。  最频繁子串：出现次数最多的子串

tags： 直接暴力怼？？

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 200005;

int n, in[28];
char s[N];
bool vis[28], used[28];
vector< int > G[N];
void print(int x)
{
    putchar('a'+x);
    vis[x] = true;
    if(G[x].size()==1)
        print(G[x][0]);
}
bool check(int x)
{
    if(vis[x]) return false;
    vis[x]=true;
    if(G[x].size()==1)
        return check(G[x][0]);
    return true;
}
int main()
{
    scanf("%d", &n);
    rep(i,1,n)
    {
        scanf("%*c%s", s+1);
        mes(vis, false);
        for(int j=1; s[j]; ++j)
        {
            int id = s[j]-'a';
            if(vis[id]) return 0*printf("NO
");
            vis[id]=true, used[id]=true;
            if(j>1) {
                if(G[s[j-1]-'a'].size()==1 && G[s[j-1]-'a'][0]==s[j]-'a')
                    continue;
                G[s[j-1]-'a'].PB(s[j]-'a');
                ++in[s[j]-'a'];
            }
            if(G[s[j-1]-'a'].size()>1 || in[s[j]-'a']>1)
                return 0*printf("NO
");
        }
    }
    mes(vis, false);
    rep(i,0,25)
        if(used[i]) {
            mes(vis, false);
            if(!check(i)) return 0*printf("NO
");
        }
    mes(vis, false);
    rep(i,0,25)
        if(used[i] && !vis[i] && in[i]==0)
            print(i);
    puts("");

    return 0;
}