1198. Substring题解

题目：

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

题意：第一行为测试用例个数，每个用例第一行为子串个数，接下来为N个子串。找出包含所有子串（每个子串仅出现一次）且字典序最小的长串并打印。

题解：由于题目限制子串个数为8以内，所以用枚举的方法是可以解的，只需存储所有结果并排序即可。但是如果是8以外呢？当N较大时，枚举肯定是不行的，时间和空间都会不够，直接将子串排序再输出是一个比较好的方法，如：a,e,b，排序后是a,b,e，输出就是正确的结果，但是直接对子串进行一半的排序也会出现问题，比如子串集为“a,b,ba”时，进行排序将会得到“a,b,ba”，而“a,ba,b”才是正确的结果。一种比较简洁的方法是重新定义对子串排序的规则，比如子串“b,ba”，要对其进行排序，先令x = b + ba = "bba", y = ba + b = "bab"，再根据组合后的x,y来判断字典序大小，此时可以看到x > y,即排序后为ba,b,这就是我们需要的结果，我利用快排来对本题进行排序，代码如下：

#include <iostream>#include <string>using namespace std;void quicksort(string* A, int start, int end);int partition(string* A, int start, int end);int main(){int test_case;cin>>test_case;while(test_case--){int n;cin>>n;string* substring_set = new string[n];for(int i = 0; i < n; i++){cin>>substring_set[i];}quicksort(substring_set, 0, n-1);for(int i = 0; i < n; i++){cout<<substring_set[i]; }cout<<endl;}return 0;}void quicksort(string* A, int start, int end){if(start < end){int q = partition(A, start, end);quicksort(A, start, q-1);quicksort(A, q+1, end);}}int partition(string* A, int start, int end){string x = A[end];int i = start - 1;for(int j = start; j < end; j++){string a,b;a = A[j] + x;b = x + A[j];if(a <= b){i += 1;string temp = A[i];A[i] = A[j];A[j] = temp;}}string temp = A[i+1];A[i+1] = A[end];A[end] = temp;return i+1;}