Substrings Sort(cf 988B)

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 $n$ ($1\le n\le 100$) — 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

Sample Input

5
a
abacaba
ba
aba
abab

Sample Output

NO

Sample Input

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3
qwerty
qwerty
qwerty

Sample Output

YES
qwerty
qwerty
qwerty

Note

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

题解：就是一个简单的排序，当时排序时没用结构体，直接用cmp里面放的strlen，没编译出来，后来验证了，有兴趣的可以试试。

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define maxn 10001
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.00000001
using namespace std;
typedef long long ll;
struct node
{
	int l;
	char b[101];
}a[101];
bool cmp(struct node x,struct node y)
{
	return x.l<y.l;
}
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
	{
		cin>>a[i].b;
		a[i].l=strlen(a[i].b);
	}
	sort(a,a+n,cmp);
	int sum=1;
	for(int i=0;i<n-1;i++)
		if(strstr(a[i+1].b,a[i].b))
			sum++;
	if(sum!=n)
		printf("NO\n");
	else
	{
		printf("YES\n");
		for(int i=0;i<n;i++)
			printf("%s\n",a[i].b);
	}
    return 0;
}