codeforce-B. Substrings Sort

最新推荐文章于 2021-11-22 22:11:32 发布

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最新推荐文章于 2021-11-22 22:11:32 发布

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本文链接：https://blog.youkuaiyun.com/SunPeishuai/article/details/81489797

本文介绍了一种算法，该算法通过对一系列字符串进行排序并检查它们之间的子串关系来确定这些字符串是否能按照特定顺序重新排列。具体实现包括输入字符串集合、按长度排序以及使用find函数检查一个字符串是否为另一个字符串的子串。

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B. Substrings Sort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn 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 aa is a substring of string bb if it is possible to choose several consecutive letters in bb in such a way that they form aa. 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 nn (1≤n≤1001≤n≤100) — the number of strings.

The next nn lines contain the given strings. The number of letters in each string is from 11 to 100100, inclusive. Each string consists of lowercase English letters.

Some strings might be equal.

Output

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

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

Examples

Input

Copy

5
a
aba
abacaba
ba
aba

Output

Copy

YES
a
ba
aba
aba
abacaba

Input

Copy

5
a
abacaba
ba
aba
abab

Output

Copy

NO

Input

Copy

3
qwerty
qwerty
qwerty

Output

Copy

YES
qwerty
qwerty
qwerty

思路：对输入的字符串按照长度进行排序，通过find函数进行查找上一个字符串是否为下一个子串，若是输出排序后的序列，否则输出NO

#include<cstdio>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
string str[108];
int cmp(string a,string  b)
{
    return a.size()<b.size();
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        cin>>str[i];
    }
    sort(str,str+n,cmp);
    for(int i=0;i<n-1;i++)
    {
        if(str[i+1].find(str[i])==str[i+1].npos)
        {
            cout<<"NO"<<endl;
            return 0;

        }
    }
    cout<<"YES"<<endl;
    for(int i=0;i<n;i++)
    {
        cout<<str[i]<<endl;
    }
    return 0;


}