1198. Substring

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Dr lee cuts a string S into N pieces,s[1],…,s[N].

Now, Dr lee gives you these N sub-strings: s[1],…s[N]. There might be several possibilities that the string S could be. For example, if Dr. lee gives you three sub-strings {“a”,“ab”,”ac”}, the string S could be “aabac”,”aacab”,”abaac”,…

Your task is to output the lexicographically smallest S.

Input

The first line of the input is a positive integer T. T is the number of the test cases followed.

The first line of each test case is a positive integer N (1 <=N<= 8 ) which represents the number of sub-strings. After that, N lines followed. The i-th line is the i-th sub-string s[i]. Assume that the length of each sub-string is positive and less than 100.

Output

The output of each test is the lexicographically smallest S. No redundant spaces are needed.

Sample Input

1
3
a
ab
ac

Sample Output

aabac

Problem Source

ZSUACM Team Member

此题由于规模不大，所以可以直接用枚举出所有可能组合，然后选择按字典序最小的字符串输出，用到递归实现了一个字符数组的全排列，每次均进行拼接保存到一个vector中用于返回，vector记得使用引用

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

void swap(string &a, string &b);
void perminent(vector<string> &result, string substrs[], int subnum, int index);

int main() {
	int t;
	cin >> t;
	while (t-- > 0) {
		int subnum;
		string substr;
		vector<string> result;
		cin >> subnum;
		string substrs[subnum];
		for (int i = 0; i < subnum; i++) {
			cin >> substr;
			substrs[i] = substr;
		}
		perminent(result, substrs, subnum, 0);

		string minstr = result[0];
		//好该死呀！！！这里把result.size()，写成了subnum！！！ 这样搞了一个多小时
		for (int i = 1; i < result.size(); i++) {
			if (minstr.compare(0, minstr.size(), result[i]) > 0) {
				minstr = result[i];
			}
		}
		cout << minstr << endl;
	}
	return 0;
}
void swap(string &a, string &b) {
	string temp = a;
	a = b;
	b = temp;
}

//获得该字符串数组的一个全排列
void perminent(vector<string> &result, string substrs[], int subnum, int index) {
	if (index == subnum-1) {
		string temp;
		for (int i = 0; i < subnum; i++) {
			temp += substrs[i];
		}
		result.push_back(temp);//result为引用调用
	} else {
		for (int i = index; i < subnum; i++) {
			//获得该数组的全排列的关键
			swap(substrs[i], substrs[index]); //把数组当前元素放在数组开头
			perminent(result, substrs, subnum, index+1);
			swap(substrs[i], substrs[index]);//还原回原来数组，继续递归，等下把数组下一个元素放在开头
		}
	}
}