CF886D Restoration of string(思维+建图)

// Problem: CF886D Restoration of string
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF886D
// Memory Limit: 250 MB
// Time Limit: 2000 ms
// Author:Cutele
// 
// Powered by CP Editor (https://cpeditor.org)

#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

inline void write(ll x){
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
    puts("");
}

#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a ;
        a = a * a ;
        b >>= 1;
    }
    return res;
}

int g[26][26],a[26],in[26],out[26],vis[26];
string res;

int dfs(int u){
	res=res+char(u+'a');
	vis[u]=1;
	rep(i,0,25)
		if(g[u][i]){
			if(vis[i]) return 0;
			return dfs(i);
		}
	return 1;
}

int main(){
	int n=read;
	rep(i,1,n){
		string s;cin>>s;
		for(int j=0;j<s.size()-1;j++){
			g[s[j]-'a'][s[j+1]-'a']=1;
		}
		if(s.size()==1) a[s[0]-'a']=1;
	}
	rep(i,0,25) rep(j,0,25)
		if(i==j&&g[i][j]){
			puts("NO");return 0;
		}
		else if(g[i][j]) in[j]++,out[i]++;
	rep(i,0,25)
		if(in[i]>1||out[i]>1){
			puts("NO");return 0;
		}
	rep(i,0,25){
		if(out[i]&&!in[i]){
			if(!dfs(i)){
				puts("NO");return 0;
			}
		}
		if(a[i]&&!out[i]&&!in[i]){
			res=res+char(i+'a');
		}
	}
	rep(i,0,25){
		if(in[i]&&out[i]&&!vis[i]){
			puts("NO");return 0;
		}
	}
	cout<<res<<endl;
	return 0;
}