Codeforces Round #486 (Div. 3)B. Substrings Sort【字符串，排序】

B. Substrings Sort

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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

Note

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

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

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题意大致就是：输入一组字符串，如果前面一个字符串是后面的子字符串则输出“YES”，并把字符串按长度依次输出，否则输出“NO”。

这道题不怎么难，但要会string类的输入输出，并且sort对string的排序是按字典序排序，所以要用到vector来排序，这道题最后的输出总是有问题，然后看了题解又改了一下，最后的输出则codeblock上总是显示语法错误，但提交上去居然过了。。。。。。。

#include<bits/stdc++.h>
using namespace std;
bool cmp(string str1,string str2){
    return str1.length()<str2.length();
}
int main(){
    //string s[101];
    int n;
    while(~scanf("%d",&n)){
        //string s[n];
        vector<string> s(n);
        //string str;
        for(int i=0;i<n;i++){
            cin>>s[i];
            //s.push_back(str);
        }
        sort(s.begin(),s.end(),cmp);
        bool flag=false;
        //string ::size_type idx;
        for(int i=0;i<n-1;i++){
            //string ::size_type idx;
            //idx=s[i+1].find(s[i]);
            if(s[i+1].find(s[i])==string::npos){
                flag=true;
                break;
            }
        }
        if(flag) cout<<"NO"<<endl;
        else{
            printf("YES\n");
            for(auto it : s) cout<<it<<endl;
        }
    }
    return 0;
}