题目链接
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.
input
5
a
aba
abacaba
ba
aba
output
YES
a
ba
aba
aba
abacaba
input
5
a
abacaba
ba
aba
abab
output
NO
input
3
qwerty
qwerty
qwerty
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： C++的string类提供了字符串中查找另一个字符串的函数find。其重载形式为：string::size_type string::find(string &);功能为在string对象中，查找参数string类型的字符串是否存在，如果存在，返回起始位置。不存在则返回 string::npos。
#include<bits/stdc++.h>
using namespace std;
int n;
string s[105];
bool cmp(string a,string b){
return a.size() < b.size();}
bool jug(int cur){
if(cur <= 0) return true;for(int i = cur - 1;i >= 0;i--){
if(s[cur].find(s[i]) == string :: npos) return false;}return true;
}
int main(){
ios :: sync_with_stdio(false);cin.tie(0);cin >> n;for(int i = 0;i < n;i++) cin >> s[i];sort(s,s + n,cmp);for(int i = n - 1;i >= 0;i--){
if(jug(i) == false){
cout << "NO" << endl;return 0;} }cout << "YES" << endl;for(int i = 0;i < n;i++) cout << s[i] << endl;return 0;
}