牛客网暑期ACM多校训练营（第十场）. Rikka with Nickname（字符串处理 + vector 容器）

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

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

复制

2
2
lubenwei
niubi
3
aa
ab
abb

输出

复制

lubenweiubi
aabb

参考代码：

#include<bits/stdc++.h>
using namespace std;
vector <int> a[28];

int main()
{
    int t,n,m;
    cin>>t;
    while(t--&&cin>>n)
    {
        for(int i=0; i<27; ++i) a[i].clear();
        int len=0;
        string str, s;
        cin >> str;
        for(int j=0; j<str.size(); j++) a[str[j]-'a'].push_back(len++);
        rep(i,2,n)
        {
            cin>>s;
            int cnt=-1,j;
            for(j=0; j<s.size(); j++)
            {
                auto it = upper_bound(a[s[j]-'a'].begin(), a[s[j]-'a'].end(), cnt);
                if(it==a[s[j]-'a'].end())
                {
                    str += s.substr(j, s.size()-j);
                    break;
                }
                cnt = *it;
            }
            for(j; j<s.size(); j++)  a[s[j]-'a'].push_back(len++);
        }
        cout << str << endl;
    }
    return 0;
}