[THUSC2016]补退选

Description

维护一个字符串集合，有三种事件，加入一个字符串，删除一个字符串，询问最早在哪个事件之后，以某个串为前缀的字符串数量超过(k)，强制在线。(n le 100000,|S| le 60),输入中的所有字符串只会包含前(10)个小写字母。

Solution

建立(Trie)树维护字符串集合，在(Trie)树上的每一个节点开一个(vector)，用来记录子树内的结束节点到达某个值的最早时刻，查询时直接找到前缀对应节点即可。

Code

#include <bits/stdc++.h>
using namespace std;

const int _ = 1e6 + 10;
int N, tim;

inline int ty() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}

inline int getstr(char *s) {
  char ch = getchar();
  int cnt = 0;
  while (ch >= 'a' && ch <= 'j') {
    s[++cnt] = ch;
    ch = getchar();
  }
  return cnt;
}

struct Trie {
  int trie[_][10], cnt[_], tot;
  vector<int> num[_];
  void insert(char *s, int len) {
    int x = 0;
    for (int i = 1; i <= len; ++i) {
      int ch = s[i] - 'a';
      if (!trie[x][ch]) trie[x][ch] = ++tot;
      x = trie[x][ch];
      ++cnt[x];
      if (num[x].size() < cnt[x]) num[x].push_back(tim);
    }
  }
  void erase(char *s, int len) {
    int x = 0;
    for (int i = 1; i <= len; ++i) {
      int ch = s[i] - 'a';
      x = trie[x][ch];
      --cnt[x];
    }
  }
  int query(char *s, int k, int len) {
    int x = 0;
    for (int i = 1; i <= len; ++i) {
      int ch = s[i] - 'a';
      if (!trie[x][ch]) return -1;
      x = trie[x][ch];
    }
    if (num[x].size() <= k)
      return -1;
    else
      return num[x][k];
  }
} tr;

int main() {
#ifndef ONLINE_JUDGE
  freopen("select.in", "r", stdin);
  freopen("select.out", "w", stdout);
#endif
  N = ty();
  int last = 0;
  for (tim = 1; tim <= N; ++tim) {
    int op = ty();
    char s[65];
    int len = getstr(s);
    if (op == 1)
      tr.insert(s, len);
    else if (op == 2)
      tr.erase(s, len);
    else if (op == 3) {
      int a = ty(), b = ty(), c = ty();
      int k = (1ll * a * abs(last) + b) % c;
      printf("%d
", last = tr.query(s, k, len));
    }
  }
  return 0;
}