Codeforces Round #486 (Div. 3) B. Substrings Sort

Posted 2020-11-13 edgsheryl

Codeforces Round #486 (Div. 3) B. Substrings Sort

题目连接：

http://codeforces.com/contest/988/problem/B

Description

You are given n strings. Each string consists of lowercase English letters. Rearrange (reorder) the given strings in such a way that for every string, all strings that are placed before it are its substrings.

String a is a substring of string b if it is possible to choose several consecutive letters in b in such a way that they form

a. For example, string "for" is contained as a substring in strings "codeforces", "for" and "therefore", but is not contained as a substring in strings "four", "fofo" and "rof".

Input

The first line contains an integer n (1≤n≤100) — the number of strings.

The next n lines contain the given strings. The number of letters in each string is from 1 to 100, inclusive. Each string consists of lowercase English letters.

Some strings might be equal.

Output

If it is impossible to reorder n given strings in required order, print "NO" (without quotes).

Otherwise print "YES" (without quotes) and n given strings in required order.

Sample Input

5
a
aba
abacaba
ba
aba

Sample Output

YES
a
ba
aba
aba
abacaba

Hint

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba"

题意

给你一堆字符串，重新排序后，前一个串是后一个串的子串，问你这样的情况存在不存在

题解：

按长度排序后，直接string.find()判断

代码

#include <bits/stdc++.h>

using namespace std;

int n;
string a[110];

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n, [](const string &str1, const string &str2) {
        return str1.size() < str2.size() || (str1.size() == str2.size() && str1 < str2);
    });
    for (int i = 1; i < n; i++) {
        int o = a[i].find(a[i - 1]);
        if (o > a[i].size() || o < 0) return 0 * puts("NO");
    }
    cout << "YES" << endl;
    for (int i = 0; i < n; i++)
        cout << a[i] << endl;
}