SCUT J.O: Substring

Substring

Time Limit:

1000MS

Memory Limit:

10000K

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

Hint

No hint.

Source

                        

                        

                        

                        

                        

                        

                        

思路：这个题比较a、b两个串的方法是比较ab和ba串的字典序，若ab<ba则a<b，反之是b<a

//#include <windows.h> //#include <iostream> //#include <fstream> #include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <algorithm> //using namespace std; //LARGE_INTEGER _lTemp; //LONGLONG _dFreq, _bTime, _eTime; //double _uTime; char str1[250], str2[250]; bool strGt(char *a, char *b){ char *ptr; ptr = str1; for (int i = 0; a[i] != '/0'; i ++) { *ptr = a[i]; ptr ++; } for (int i = 0; b[i] != '/0'; i ++) { *ptr = b[i]; ptr ++; } *ptr = '/0'; ptr = str2; for (int i = 0; b[i] != '/0'; i ++) { *ptr = b[i]; ptr ++; } for (int i = 0; a[i] != '/0'; i ++) { *ptr = a[i]; ptr ++; } *ptr = '/0'; for (int i = 0; str1[i] != '/0'; i ++) if (str1[i] > str2[i]) return true; else if (str1[i] < str2[i]) return false; return false; } int main(){ // freopen("in.txt", "r", stdin); //重定向输入 // freopen("out.txt", "w", stdout);//重定向输出 // QueryPerformanceFrequency(&_lTemp); //获取运算频率 // _dFreq = _lTemp.QuadPart; //获取运算频率大整数 // QueryPerformanceCounter(&_lTemp); //获取开始时间 // _bTime = _lTemp.QuadPart; //获取开始时间大整数 int n, m, ord[10]; char list[10][120]; scanf("%d", &n); while (n --){ scanf("%d", &m); for (int i = 0; i < m; i ++){ scanf("%s", list[i]); ord[i] = i; } for (int i = 0; i < m; i ++){ int min = i; for (int j = i + 1; j < m; j ++) if (strGt(list[ord[min]], list[ord[j]])) min = j; int tmp = ord[min]; ord[min] = ord[i]; ord[i] = tmp; } for (int i = 0; i < m; i ++) printf("%s", list[ord[i]]); printf("/n"); } // QueryPerformanceCounter(&_lTemp); //获取终止时间 // _eTime = _lTemp.QuadPart; //获取终止时间大整数 // _uTime = 1000 * double(_eTime - _bTime) / double(_dFreq); //计算总耗时，转化成ms // printf("Time used: %012.3f ms/n", _uTime); //输出总耗时，单位是ms // fclose(stdin); //关闭文件 // fclose(stdout); //关闭文件 return 0; }