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".
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.
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.
5 a aba abacaba ba aba
YES a ba aba aba abacaba
5 a abacaba ba aba abab
NO
3 qwerty qwerty qwerty
YES qwerty qwerty qwerty
In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".
题意:先按长度排列字符串,然后从小到大前一项是后一项的子串,直到最后一个
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
string str[110];
int n;
bool cmp(string a,string b)
{
return a.size()<b.size();
}
int main()
{
bool flag=true;
cin>>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])==string::npos)//找不到子串
flag=false;
if(!flag) puts("NO");
else
{
puts("YES");
for(int i=0;i<n;i++)
cout<<str[i]<<endl;
}
return 0;
}