[ARC081E] Don't Be a Subsequence

考虑建出子序列自动机，然后在上面广搜即可。因为是广搜，所以可以保证找出的是最短的符合条件的字符串。注意广搜时要判 vis。复杂度线性。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

const int N = 5e5 + 5;
string s;
int n;
int son[N][26], nxt[26];
int pre[N];
bool vis[N];

void bfs()
{
	queue<pair<int, string> > q;
	q.push(make_pair(0, ""));
	while (q.size())
	{
		if (vis[q.front().first])
		{
			q.pop();
			continue;
		}
		vis[q.front().first] = 1;
		for (int i = 0; i < 26; i++)
		{
			q.front().second += (i + 'a');
			if (!son[q.front().first][i])
			{
				cout << q.front().second << "\n";
				return;
			}
			q.push(make_pair(son[q.front().first][i], q.front().second));
			q.front().second.pop_back();
		}
		q.pop();
	}
	//cout << vv.size() << "\n";
}

int main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	cin >> s;
	n = s.size();
	s = (string)" " + s;
	for (int i = 0; i < 26; i++) nxt[i] = 1;
	for (int i = 0; i <= n; i++)
	{
		for (int j = 0; j < 26; j++)
		{
			nxt[j] = max(nxt[j], i + 1);
			for (int k = nxt[j]; k <= n; k++)
			{
				nxt[j] = k;
				if (s[k] - 'a' == j)
				{
					son[i][j] = k;
					break;
				}
			}
		}
	}
	bfs();
	return 0;
}