有nn个事件，每次给出SS：

  1. 在当前字符串集中加入SS
  2. 在当前字符串集中删去SS
  3. 查询最早在哪个事件之后，SS为前缀的字符串数量超过了vv

强制在线

n≤105,∣S∣≤60n\le 10^5, |S|\le 60

LOJ2291

Solution

一开始还以为要二分答案，后面发现直接在Trie上模拟，对于每个节点维护一个vector暴力记录答案即可

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> T read ()
{
    T sum = 0, fl = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
    for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
    return sum * fl;
}

inline void proc_status()
{
    ifstream t ("/proc/self/status");
    cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) <<endl;
}

const int Maxn = 1e5 + 100;

int N, M;

int node_cnt = 1;
struct info
{
    int ch[27], cnt; 
	vector <int> p;
}node[Maxn * 40];

struct TRIE
{

    inline void insert (char *S, int op, int id)
    {
        int n = strlen(S), now= 1;
        for (int i = 0; i < n; ++i)
        {
			int x = S[i] - 'a';
			if (!node[now].ch[x]) node[now].ch[x] = ++node_cnt;
			now = node[now].ch[x];
			node[now].cnt += op;
			if (node[now].cnt > node[now].p.size()) node[now].p.pb(id);
        }
    }

	inline int query (char *S, int times)
	{
		int n = strlen(S), now = 1;
		for (int i = 0; i < n; ++i) now = node[now].ch[S[i] - 'a'];
		if (!now || node[now].p.size() <= times) return -1;
		return node[now].p[times];
	}

} T;

inline void Solve ()
{
	int ans = 0;
	for (int id = 1; id <= M; ++id)
	{
		int op = read<int>(); 
		char S[100]; scanf("%s", S);

		if (op <= 2) T.insert(S, op == 1 ? 1 : -1, id);
		else
		{
			int a = read<int>(), b = read<int>(), c = read<int>();
			int sum = ((LL)a * abs(ans) + b) % c;
			ans = T.query (S, sum);
			printf("%d\n", ans);
		}
	}
}

inline void Input ()
{
	M = read<int>();
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("A.in", "r", stdin);
    freopen("A.out", "w", stdout);
#endif
    Input();
    Solve();
    return 0;
}