[NOI2017]蚯蚓排队

嘟嘟嘟


现在看来这道题还不是特别难。
别一看到字符串就想SAM


看到\(k\)很小，所以我们可以搞一个单次修改复杂度跟\(k\)有关的算法。
能想到，每一次断开或链接，最多只会影响\(k ^ 2\)个长度为\(k\)的区间。所以我们开一个哈希表，每一次拼接时就往哈希表里加入\(k ^ 2\)个新的哈希值，断链的时候就把这些哈希值减去。然后查询的时候只要扫一遍\(s\)，每遇到长度为\(k\)的子串就再查一下。
具体的操作要用到链表，对于每一个节点分别维护一个pre和suf指针即可。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const ll mod = 998244353;
const int maxn = 2e5 + 5;
const int maxm = 2e7 + 5;
const int LIM = 50;
const ull NUM = 233;
const ull P = 19260817;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
  freopen("worm.in", "r", stdin);
  freopen("worm.out", "w", stdout);
#endif
}

char s[maxm];
int n, m, a[maxn];
struct Hash
{
  int head[maxm], nxt[maxm], val[maxm], tot;
  ull to[maxm];
  In void update(ull x, int sum)
  {
    int h = x % P;
    for(int i = head[h]; i; i = nxt[i])
      if(to[i] == x) {val[i] += sum; return;}
    nxt[++tot] = head[h], to[tot] = x, val[tot] = sum, head[h] = tot;
  }
  In int query(ull x)
  {
    for(int i = head[x % P]; i; i = nxt[i])
      if(to[i] == x) return val[i];
    return 0;
  }
}H;

ull p[maxn];
int pre[maxn], suf[maxn];
In void merge(int x, int y)
{
  ull now = 0;
  pre[y] = x, suf[x] = y;
  for(int i = x, l1 = 1; i && l1 <= LIM; i = pre[i], ++l1)
    {
      now += p[l1 - 1] * a[i]; ull tp = now;
      for(int j = y, l2 = l1 + 1; j && l2 <= LIM; j = suf[j], ++l2)
	{
	  tp = tp * NUM + a[j];
	  H.update(tp, 1);
	}
    }
}
In void split(int x)
{
  int y = suf[x]; ull now = 0;
  for(int i = x, l1 = 1; i && l1 <= LIM; i = pre[i], ++l1)
    {
      now += p[l1 - 1] * a[i]; ull tp = now;
      for(int j = y, l2 = l1 + 1; j && l2 <= LIM; j = suf[j], ++l2)
	{
	  tp = tp * NUM + a[j];
	  H.update(tp, -1);
	}
    }
  suf[x] = 0, pre[y] = 0;
}
In ll query(const int& k)
{
  ull now = 0, ret = 1;
  int len = strlen(s);
  for(int i = 0; i < len; ++i)   //滚动哈希
    {
      now = now * NUM + (s[i] - '0');
      if(i >= k - 1)
	{
	  ret = ret * H.query(now) % mod;
	  now -= p[k - 1] * (s[i - k + 1] - '0');
	}
    }
  return ret;
}

int main()
{
  MYFILE();
  n = read(), m = read();
  for(int i = 1; i <= n; ++i) a[i] = read(), H.update(a[i], 1);
  p[0] = 1;
  for(int i = 1; i <= LIM; ++i) p[i] = p[i - 1] * NUM;
  for(int i = 1; i <= m; ++i)
    {
      int op = read();
      if(op == 1)
	{
	  int x = read(), y = read();
	  merge(x, y);
	}
      else if(op == 2)
	{
	  int x = read();
	  split(x);
	}
      else
	{
	  scanf("%s", s); int k = read();
	  write(query(k)), enter;
	}
    }
  return 0;
}