Educational Codeforces Round 16 F - String Set Queries (CDQ分治+AC自动机)
题目链接:
http://codeforces.com/contest/710/problem/F
题意:
1. 集合加入一个串
2. 集合删除一个串
3. 问集合中的串在给定串中出现的次数和(可重复)
要求强制在线
思路:
如果是离线的话,直接cdq分治,每次求左半边的AC自动机去更新右半边
强制在线的话,只要保存一下log个左半边的AC自动机就好了
复杂度:O((n+s)log(n))
#include<bits/stdc++.h> using namespace std; const int N=900005; int ch[N][26],fail[N],cnt[N],f[N]; int tot; int newnode(){ tot++; for (int i=0;i<26;i++) ch[tot][i]=0; fail[tot]=0; cnt[tot]=0; return tot; } int merge(int x,int y){ if (!x||!y) return x+y; cnt[x]+=cnt[y]; for (int i=0;i<26;i++) ch[x][i]=merge(ch[x][i],ch[y][i]); return x; } int root[20]; void insert(int root,char *s,int y){ int now=root; for (int i=0;s[i];i++){ int w=s[i]-'a'; if (!ch[now][w]) ch[now][w]=newnode(); now=ch[now][w]; } cnt[now]+=y; } int go(int root,int u,int w){ while(u!=root&&!ch[u][w]) u=fail[u]; if (ch[u][w]) return ch[u][w]; return u; } void build(int root){ queue<int>Q; fail[root]=root; for (int i=0;i<26;i++){ int v=ch[root][i]; if (v){ fail[v]=root; Q.push(v); } } while(!Q.empty()){ int u=Q.front(); Q.pop(); f[u]=f[fail[u]]+cnt[u]; for (int i=0;i<26;i++){ int v=ch[u][i]; if (v){ fail[v]=go(root,fail[u],i); Q.push(v); } } } } int query(int root,char *s){ int ret=0; int now=root; for (int i=0;s[i];i++){ int w=s[i]-'a'; now=go(root,now,w); ret+=f[now]; } return ret; } void init(int &root){ root=newnode(); build(root); } int tp; char s[N]; int solve(int l,int r,int deep){ if (l==r){ cin>>tp>>s; init(root[deep]); if (tp<3){ int y; if (tp==1) y=1; else y=-1; insert(root[deep],s,y); } else{ long long ans=0; for (int d=1;d<deep;d++) ans+=query(root[d],s); cout<<ans<<endl; fflush(stdout); } return root[deep]; } init(root[deep]); int mid=(l+r)/2; int rtl=solve(l,mid,deep+1); root[deep]=merge(root[deep],rtl); build(root[deep]); int rtr=solve(mid+1,r,deep+1); root[deep]=merge(root[deep],rtr); build(root[deep]); return root[deep]; } int main(){ ios::sync_with_stdio(false); int n; cin>>n; solve(1,n,1); }
2022年9月05日 23:13
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