Codeforces 890 D. Restoration of string （技巧）

Description

A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.

You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print “NO” (without quotes).

A substring of a string is a contiguous subsequence of letters in the string. For example, “ab”, “c”, “abc” are substrings of string “abc”, while “ac” is not a substring of that string.

The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.

String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.

Input

The first line contains integer n (1 ≤ n ≤ 10^5) — the number of strings in the set.

Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.

The total length of the strings doesn’t exceed 10^5.

Output

Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print “NO” (without quotes) if there are no good strings.

Examples input

4
mail
ai
lru
cf

Examples output

cfmailru

题意

构造一个字典序最小的字符串，满足给定的所有串都是它的子串且这些串的出现频率最高。

思路

显然，给定的一个串中如果出现两个相同的字母结果一定是

NO

。

对于

mail

与

ai

，

mail

与

ailkk

，

mail

与

kkmai

这三种类型的子串我们采取合并措施，若出现比如

abc

与

abxc

这种部分不连续的匹配或匹配错误则一定输出

NO

。

暴力枚举即可，因为相同字母已经把每个串的长度限制在了 26 以内，再加上部分不连续匹配的限制条件，最后能对答案做出贡献的串很少很少，想必测试数据中 n 在某个值以上的结果都是

NO

。

合并完的串之间是没有任何关联的，我们可以对这些串排序输出即字典序最小解。

AC 代码

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
cin.tie(0);\
cout.tie(0);
typedef long long LL;
using namespace std;
const int maxn = 2e6+10;

string s[maxn];
int vis[26];
int n;

bool solve2(int s1,int s2)
{
int len1 = s[s1].length();
int len2 = s[s2].length();
memset(vis,0,sizeof(vis));
for(int i=0; i<len1; i++)
vis[s[s1][i]-'a']=i+1;
for(int i=0; i<len2; i++)
if(vis[s[s2][i]-'a'])   //找的第一个相同位
{
int yi = vis[s[s2][i]-'a']-1;
int ni = i;
string cnt;
for(int j=-26; j<26; j++)   //合并两个串
{
if(yi+j>=0&&yi+j<len1&&ni+j>=0&&ni+j<len2)
{
if(s[s1][yi+j]!=s[s2][ni+j])
return false;
cnt+=s[s1][yi+j];
}
else if(yi+j>=0&&yi+j<len1)
cnt+=s[s1][yi+j];
else if(ni+j>=0&&ni+j<len2)
cnt+=s[s2][ni+j];
}
memset(vis,false,sizeof(vis));  //合并完以后检查是否合法
for(int j=0; j<(int)cnt.length(); j++)
if(++vis[cnt[j]-'a']>1)
return false;
s[n++] = cnt;
s[s1] = "";
s[s2] = "";
return true;
}
return true;
}

void solve()
{
for(int i=0; i<n; i++)
{
memset(vis,0,sizeof(vis));
for(auto j:s[i])
if(++vis[j-'a']>1)
{
cout<<"NO"<<endl;
return;
}
}
for(int i=0; i<n; i++)
for(int j=i+1; j<n; j++)
if(s[i]!=""&&s[j]!=""&&!solve2(i,j))
{
cout<<"NO"<<endl;
return;
}
sort(s,s+n);
for(int i=0; i<n; i++)
cout<<s[i];
}

int main()
{
IO;
cin>>n;
for(int i=0; i<n; i++)
cin>>s[i];
solve();
return 0;
}