Codeforces Round #445 (Div. 2): D. Restoration of string（模拟）

B. Restoration of string

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ≤ 105) — 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 105.

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

4mailailrucf

output

cfmailru

input

3kekpreceqcheburek

output

NO

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平凡子串，如果不存在输出NO

最平凡子串：出现次数最多的子串

思路：

①很显然如果某个子串中有两个相同的字符，答案一定是NO

②如果两个不同的子串中有相同的字符

那么这两个子串一定可以通过拼在一起使得每个字符都只出现1次（也就是一个串的前缀和另一个串的后缀相同）

如果不能拼也是NO

这样思路就明显了：对于每个串先检查是否每个字符都独一无二

然后再暴力所有字符串看是否能和其中一个拼起来

注意拼起来之后要继续拼！直到对于所有字符串所有字符都只出现一次

中间出现任何问题都是NO

代码写丑了

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int cnt[33];char str[100005], ans[60][60];typedef struct Res{char str[60];bool operator < (const Res &b) const{if(strcmp(str+1, b.str+1)<0)return 1;return 0;}}Word;Word s[60];int main(void){int n, i, j, k, p, q, len;scanf("%d", &n);k = 0;for(i=1;i<=n;i++){scanf("%s", str+1);len = strlen(str+1);if(len>=27){printf("NO\n");return 0;}loop:for(j=1;j<=k;j++){if(strchr(ans[j]+1, str[len])!=NULL && strchr(ans[j]+1, str[1])==NULL){for(p=1;p<=56;p++)swap(ans[j][p], str[p]);len = strlen(str+1);if(strchr(ans[j]+1, str[1])==NULL){printf("NO\n");return 0;}}if(strchr(ans[j]+1, str[1])!=NULL){for(p=1;ans[j][p]!=str[1];p++);for(q=1;q<=len;q++){if(ans[j][p+q-1]!=str[q] && ans[j][p+q-1]!=0){printf("NO\n");return 0;}ans[j][p+q-1] = str[q];}memset(str, 0, sizeof(str));break;}}if(j!=k+1){for(p=1;p<=56;p++)swap(ans[j][p], ans[k][p]);for(p=1;p<=56;p++)swap(ans[k][p], str[p]);len = strlen(str+1);k--;goto loop;}if(j==k+1){k++;for(p=1;p<=56;p++)swap(ans[k][p], str[p]);j = k;}memset(cnt, 0, sizeof(cnt));for(p=1;p<=len;p++){if(cnt[ans[j][p]-'a']==1){printf("NO\n");return 0;}cnt[ans[j][p]-'a'] = 1;}}memset(cnt, 0, sizeof(cnt));for(i=1;i<=k;i++){for(j=1;ans[i][j]!='\0';j++)s[i].str[j] = ans[i][j];for(j=1;s[i].str[j]!='\0';j++){if(cnt[s[i].str[j]-'a']==1){printf("NO\n");return 0;}cnt[s[i].str[j]-'a'] = 1;}}sort(s+1, s+k+1);for(i=1;i<=k;i++)printf("%s", s[i].str+1);puts("");return 0;}/*3swtnmzzs*/