链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
来源：牛客网
Rikka with Nickname
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.)
链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
来源：牛客网
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:
链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
来源：牛客网
输出描述:
For each testcase, output a single line with a single string, the result nickname.
链接：https://ac.nowcoder.com/acm/contest/148/J?&headNav=acm
来源：牛客网
示例1
输入
复制
2
2
lubenwei
niubi
3
aa
ab
abb
输出
复制
lubenweiubi
aabb
题意：
思路：
维护一个vector
对于每一个新尝试加入的字符，我们去vector中二分查找尽可能靠前的位置，如果找不到就只能老老实实的加入到ans中。
能找到的话，把位置赋值为last，在 下一次查找中使用。
细节见代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
std::vector<int> v[50];
string ans, str;
int n;
int main()
{
//freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
//freopen("D:\\common_text\code_stream\\out.txt","w",stdout);
int t;
gbtb;
cin >> t;
while (t--)
{
// MS0(f);
for (char i = 'a'; i <= 'z'; ++i)
{
v[i - 'a'].clear();
}
cin >> n;
ans = "";
int last = -1;
cin >> ans;
rep(i, 0, sz(ans))
{
v[ans[i] - 'a'].push_back(i);
}
repd(i, 2, n)
{
cin >> str;
int len = str.length();
last = -1;
repd(j, 0, len - 1)
{
if (sz(v[str[j] - 'a']))
{
int id = lower_bound(ALL(v[str[j] - 'a']), last) - v[str[j] - 'a'].begin();
if (id == sz(v[str[j] - 'a']))
{
ans.push_back(str[j]);
v[str[j] - 'a'].push_back(sz(ans) - 1);
last = inf;
} else
{
last = v[str[j] - 'a'][id] + 1;
}
} else
{
ans.push_back(str[j]);
v[str[j] - 'a'].push_back(sz(ans) - 1);
last = inf;
}
}
}
cout << ans << endl;
}
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
}
else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}