2017 青岛现场赛 Suffix

Consider n given non-empty strings denoted by s1 , s2 , · · · , sn . Now for each of them, you need to select a corresponding suffix, denoted by suf1, suf2, · · · , sufn. For each string si, the suffix sufi is a non-empty substring whose right endpoint is the endpoint of the entire string. For instance, all suffixes of the string “jiangsu” are “u”, “su”, “gsu”, “ngsu”, “angsu”, “iangsu” and itself.

All selected suffixes could assemble into a long string T = $suf1suf_1suf1 + suf2suf_2suf2 + · · · + sufnsuf_nsufn . Here plus signs indicate additions of strings placing the latter at the tail of the former. Your selections of suffixes would determine the lexicographical order of T . Now, your mission is to find the one with minimum lexicographical order.$

Here is a hint about lexicographical order. To compare strings of different lengths, the shorter string is usually padded at the end with enough “blanks” which is a special symbol that is treated as smaller than every letters.

Input

The first line of input contains an integer T which is the total number of test cases. For each case, the first line contains an positive integer n. Each of the following n lines contains a string entirely in lowercase, corresponding to $s1s_1s1 , s2s_2s2 , · · · , sns_nsn . The summation of lengths of all strings in input is smaller or equal to 500000.$

Output

For each test case, output the string T with minimum lexicographical order.

样例输入

3
3
bbb
aaa
ccc
3
aba
aab
bab
2
abababbaabbababba
abbabbabbbababbab

样例输出

baaac
aaabab
aab

题意：给n个字符串，从每个字符串种选择一个最小后缀，要求把每个字符串的最小后缀拼接在一起组成一个字典序最小的字符串并输出

题解：暴力，从最后一个字符串开始处理，用for找到每个字符串的最小后缀，再用s.substr()函数拼接后缀，输出结果

注意：如果每次找到一个字符串的最小后缀，然后把所用后缀拼接起来可能并不是最终答案
比如输入2个字符串
aabaa
cc
第一个字符串的最小后缀是aa
第二个最小后缀是c
如果简单拼接，答案是aac
但真正结果是aabaac
因为后面的答案会影响前面的后缀串大小，因此我们得先求出最后的最小后缀串，然后将其加到上一个串的结尾，再求前面的串的最小的后缀串，以此类推再上一个串即可

题目来源

ACM-ICPC 2017 Asia Qingdao

/*
string s;
s.substr(start,len);//字符串提取函数，从位置start开始提取长度为len的子串
*/

#include<iostream>
#include<stdio.h>
#include<string>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int  N=500005;
string str[200005];
int anslen,nowlen;
int len[N];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        str[0]="";
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            cin>>str[i];
            len[i]=str[i].size();
        }
 
        anslen=1;
        nowlen=1;
 
        for(int i=n; i>=1; i--)//从最后一个字符串开始处理
        {
            int pos=1;
            nowlen=str[i].size();
            for(int j=1; j<=len[i]; j++)//j<=len[i]保证处理每一个字符串的时候都会取一个后缀
            {
                if(str[i].substr(pos-1,nowlen-pos+1)>str[i].substr(j-1,nowlen-j+1) )//找到后缀最小的字符串开始的位置pos
                {
                    pos=j;
                }
            }
            str[i-1]+=str[i].substr(pos-1,nowlen-pos+1);//将第i个字符串的最小后缀拼接到第i-1个字符串上
        }
        cout<<str[0]<<endl;
 
 
    }
 
    return 0;
}