Rikka with Nickname （简单题）

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

输入

2
2
lubenwei
niubi
3
aa
ab
abb

输出

lubenweiubi
aabb

题解：水题，一开始做的时候用了string的find、erase就TLE了，把这些换成普通的搜索时间竟然大大减少了

下面第一份是超时的代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const ll mod=998244353;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
#define MAX 100005
int T;
int n;
string s,ans,t,ss;
int main()
{
      cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    cin>>T;
    while(T--)
    {
        cin>>n;
        cin>>s;
        ans=s;
        for(int d=0;d<n-1;d++)
        {
            t=ans;
            cin>>s;
            int k=0;
            for(int i=0;i<s.size();i++)
            {
                int tt=t.find(s[i]);
                if(tt>=0)
                {
                    k++;
                    tt++;
                    t.erase(0,tt);
                }
                else break;
            }
            for(int j=k;j<s.size();j++)
                ans+=s[j];
        }
        cout<<ans<<endl;
    }
    return 0;
}

这是AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const ll mod=998244353;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
#define MAX 100005
int T;
int n;
string s,ans,t,ss;
int main()
{
      cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);
    cin>>T;
    while(T--)
    {
        cin>>n;
        cin>>s;
        ans=s;
        for(int d=0;d<n-1;d++)
        {
            t=ans;
            cin>>s;
            int k=0;
            int f=0;
          while(k<t.size())
          {
              if(s[f]==t[k])f++,k++;
              else k++;
              if(f==s.size())
                break;
          }
            for(int j=f;j<s.size();j++)
                ans+=s[j];
        }
        cout<<ans<<endl;
    }
    return 0;
}