牛客网多校第10场J-Rikka with Nickname

链接：https://www.nowcoder.com/acm/contest/148/J
来源：牛客网

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

Sometimes you may want to write a sentence into your nickname like "lubenwei niubi". But how to change it into a single word? Connect them one by one like "lubenweiniubi" looks stupid.

To generate a better nickname, Rikka designs a non-trivial algorithm to merge a string sequence s1...sn into a single string. The algorithm starts with s=s1 and merges s2...sn into s one by one. The result of merging t into s is the shortest string r which satisfies s is a prefix of r and t is a subsequence of r.(If there are still multiple candidates, take the lexicographic order smallest one.)

String s is a prefix of r if and only if |s| ≤ |r| and for all index i ∈ [1, |s|], si = ri.

String s is a subsequence of r if and only if there is an index sequence which satisfies .

For example, if we want to generate a nickname from "lubenwei niubi", we will merge "niubi" into "lubenwei", and the result is "lubenweiubi".

Now, given a sentence s1...sn with n words, Rikka wants you to calculate the resulting nickname generated by this algorithm.

输入描述:

The first line contains a single number t(1 ≤ t ≤ 3), the number of testcases.

For each testcase, the first line contains one single integer n(1 ≤ n ≤ 106).

Then n lines follow, each line contains a lowercase string .

输出描述:

For each testcase, output a single line with a single string, the result nickname.

 

示例1

输入

2
2
lubenwei
niubi
3
aa
ab
abb

输出

lubenweiubi
aabb

以下代码均可AC，啊啊啊，看错了，不是程序跑了7s，是提交用时。。。我就说怎么2秒的限时跑7s还能AC

好烦，调程序调了好久。想麻烦了，但是我的程序也能过，虽然跑了7s

死在了j+c-cnt上，忘记减掉cnt了qwq

耗时268ms

#include
#define ll long long
using namespace std;
vector  vis[233];
int main()
{
    int t,n;
    scanf("%d",&t);

    while(t--)
    {
        scanf("%d",&n);
        string a[n+10];
        for(int i=0;i>a[i];
        string s = a[0];

        for(int i=0;i

 

然后想着这样写会不会减小复杂度然鹅却是，402 ms

#include
#define ll long long
using namespace std;
vector  vis[233];
int main()
{
    int t,n;
    scanf("%d",&t);

    while(t--)
    {
        scanf("%d",&n);
        string a[n+10];
        for(int i=0;i>a[i];
            if(i)
            {
                if(!vis[a[i][0]].size())
                {
                    int c = a[0].size();
                    for(int j=0;j

 

然后我觉得c++的string耗时应该比较大，然后改成了下面这个，耗时154 ms

 

#include
using namespace std;
char a[1000005];
char b[1000005];
vector  vis[233];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        scanf("%s",a);
        for(int i=0;a[i];i++)
            vis[a[i]].push_back(i);
        for(int i=0;i

 

最后，一种简单的写法（纯暴力，可怕可怕），耗时183ms

 

#include
using namespace std;
typedef long long ll;

char a[1000005];
char b[1000005];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        scanf("%s",a);
        n--;
        while(n--)
        {
            scanf("%s",b);
            int pos = 0;
            int len = strlen(b);
            for(int i=0;pos