[BZOJ4943][NOI2017]蚯蚓

bzoj
luogu
uoj

sol

用链表维护蚯蚓的相对顺序。
发现每次询问的\(k\le50\)，所以在每次链表的合并与分离时，暴力找前后\(50\)个位置，然后\(O(k^2)\)地塞进\(Hash\ Table\)或者从\(Hash\ Table\)里面删除。
因为分离操作最多\(10^3\)次，所以合并操作最多\(n+999\)次，复杂度显然会是对的。

code

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int gi(){
	int x=0,w=1;char ch=getchar();
	while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if (ch=='-') w=0,ch=getchar();
	while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
#define ull unsigned long long
const int N = 2e5+5;
const int base = 20020415;
const int K = 51;
const int mod = 998244353;
int n,m,a[N],pre[N],nxt[N],f[N];ull pw[N],g[N];char s[N*50];
struct edge{ull v;int nxt,w;}E[N*99];int head[4194304],cnt;
void add(ull val,int op){
	int u=val&4194303,e=head[u];
	for (;e;e=E[e].nxt) if (E[e].v==val) {E[e].w+=op;return;}
	E[e=++cnt]=(edge){val,head[u],op};head[u]=e;
}
int query(ull val){
	int u=val&4194303,e=head[u];
	for (;e;e=E[e].nxt) if (E[e].v==val) return E[e].w;
	return 0;
}
void merge(){
	int x=gi(),y=gi(),l=K,r=K-1;
	for (int i=x;i&&l>1;i=pre[i]) f[--l]=a[i];
	for (int i=y;i&&r<K*2;i=nxt[i]) f[++r]=a[i];
	for (int i=1;i<=r;++i) g[i]=g[i-1]*base+f[i];
	for (int i=l;i<K;++i)
		for (int j=K;j<=min(r,i+49);++j)
			add(g[j]-g[i-1]*pw[j-i+1],1);
	nxt[x]=y;pre[y]=x;
}
void split(){
	int x=gi(),y=nxt[x],l=K,r=K-1;
	for (int i=x;i&&l>1;i=pre[i]) f[--l]=a[i];
	for (int i=y;i&&r<K*2;i=nxt[i]) f[++r]=a[i];
	for (int i=1;i<=r;++i) g[i]=g[i-1]*base+f[i];
	for (int i=l;i<K;++i)
		for (int j=K;j<=min(r,i+49);++j)
			add(g[j]-g[i-1]*pw[j-i+1],-1);
	nxt[x]=pre[y]=0;
}
int query(){
	scanf("%s",s+1);int k=gi(),len=strlen(s+1);ull val=0;
	for (int i=1;i<=k;++i) val=val*base+s[i];
	int res=query(val);
	for (int i=k+1;i<=len;++i){
		val=val*base+s[i];
		val-=s[i-k]*pw[k];
		res=1ll*res*query(val)%mod;
	}
	return res;
}
int main(){
	n=gi();m=gi();pw[0]=1;
	for (int i=1;i<N;++i) pw[i]=pw[i-1]*base;
	for (int i=1;i<=n;++i) add(a[i]=gi()+'0',1);
	while (m--){
		int op=gi();
		if (op==1) merge();
		else if (op==2) split();
		else printf("%d\n",query());
	}
	return 0;
}