Codeforces Round #486 (Div. 3) B题

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.

想法

该题的意思就是给你n个字符串，如果这n个字符串能够满足：后出现的串必须有一个子串是前一个串。

那意思就是说后边的字符串一定不小于前边的字符串。

那我们可以先排个序，然后在判断后边的字符串是否包含前边的，如果不包含就是return；反之，继续判断下一个。

我写的时候用到了pair对，first存放字符串的长度(便于用sort排序)，second存放字符串。

判断后边是否包含前边的。我用的是find，如果查找成功就返回查找到的第一个位置，否则返回-1。

代码

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;

int main(){
    int n;
    cin>>n;
    getchar();
    pair<int,string> p[n];
    for(int i=0;i<n;i++){
        getline(cin,p[i].second);
        p[i].first = (p[i].second).length();
    }
    sort(p,p+n);
    for(int i=0;i<n-1;i++){
        string a = p[i].second;
        string b = p[i+1].second;
        if(b.find(a) != -1){
            continue;
        }else{
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    for(int i=0;i<n;i++){
        cout<<p[i].second<<endl;
    }
    return 0;
}

配上cf大佬的代码

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n;
	cin >> n;
	vector<string> s(n);
	for (int i = 0; i < n; ++i)
		cin >> s[i];
		
	sort(s.begin(), s.end(), [&] (const string &s, const string &t) {
		return s.size() < t.size();
	});
	
	for (int i = 0; i < n - 1; ++i) {
		if (s[i + 1].find(s[i]) == string::npos) {
			cout << "NO\n";
			return 0;
		}
	}
	
	cout << "YES\n";
	for (auto it : s)
		cout << it << endl;
	
	return 0;
}