洛谷3823 [NOI2017] 蚯蚓排队 【哈希】

题目分析:

从$\sum|S|$入手。共考虑$\sum|S|$个$f(t)$。所以我们要一个对于每个$f(t)$在$O(1)$求解的算法。不难想到是哈希。

然后考虑分裂和合并操作。一次合并操作要考虑合并点之前的$O(k)$个点向后衔接的哈希值。共$O(k^2)$。看似超时实则不然。一个串最多$O(nk)$个哈希结果，所以均摊入手。

对于分裂和重合并不能均摊，所以多了一个$O(ck^2)$。

这题的合并和分裂只和合并点和分裂点有关，而这个信息是给出的，所以不需要额外使用数据结构维护结果。

时间复杂度$O(nk+mk+ck^2+\sum |s|)$

代码：

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 using namespace std;

 const int maxsz = ;
 const int maxn = ;
 const int mod = ;
 const int hmod1 = ;
 const int hmod2 = ;

 int n,m,maxx;
 int a[maxn],pts[maxn];
 int ans[maxn],Num,nouse[],nuse[];
 int hsh[hmod1+],Nxt[maxn*],real[maxn*],val[maxn*],hnum;

 struct query{ int cas;string str; int l,r; }Q[maxn];

 int global = ;

 struct linktable{int pre[maxn>>],nxt[maxn>>];}T;

 void in(int &x){
     char ch = getchar();
     while(ch > '' || ch < '') ch = getchar();
     while(ch <= '' && ch >= '') x = x*+ch-'',ch = getchar();
 }

 void readstring(string &str){
     char ch = getchar();
     while(ch > '' || ch < '') ch =getchar();
     while(ch <= '' && ch >= '') str.push_back(ch),ch=getchar();
 }

 void read(){
     in(n),in(m);
     for(int i=;i<=n;i++) {in(a[i]);}
     for(int i=;i<=m;i++){
     in(Q[i].cas);
     if(Q[i].cas == ) in(Q[i].l),in(Q[i].r);
     else if(Q[i].cas == ) in(Q[i].l);
     else {readstring(Q[i].str),in(Q[i].l);Q[i].r = ++Num;maxx = max(maxx,Q[i].l);}
     }
 }

 int imnum[],numnum;

 void Add_hash(int a1,int a2,int dr){
     int p1 = hsh[a1];
     while(true) {
     if(real[p1] == a2) {val[p1]+=dr;return;}
     if(Nxt[p1]) p1 = Nxt[p1];
     else break;
     }
     hnum++; if(p1) Nxt[p1] = hnum;
     p1 = hnum; val[p1] = ; real[p1] = a2;
     if(!hsh[a1]) hsh[a1] = p1;
 }

 void buildnew(int lft,int rgt,int dr){
     numnum = ;int now = lft;
     while(true){
     if(numnum == maxx-) break;
     imnum[++numnum] = a[now];
     if(T.pre[now]) now = T.pre[now];
     else break;
     }
     now = rgt;long long um = ,vm = ;
     for(int i=;i<=numnum;i++){
     um = (1ll*nouse[i-]*imnum[i]+um)%hmod1;
     vm = (1ll*nuse[i-]*imnum[i]+vm)%hmod2;
     long long r1 = um,r2 = vm;
     for(int j=i+,jj=rgt;j<=maxx;j++){
         r1 = (r1*+a[jj])%hmod1; r2 = (r2*+a[jj])%hmod2;
         Add_hash(r1,r2,dr); if(!T.nxt[jj])break; jj = T.nxt[jj];
         global++;
     }
     }
 }

 void link(int lft,int rgt){
     buildnew(lft,rgt,);
     T.nxt[lft] = rgt; T.pre[rgt] = lft;
 }

 void cut(int place){
     buildnew(place,T.nxt[place],-);
     T.pre[T.nxt[place]] = ; T.nxt[place] = ;
 }

 int COUNT(int alpha,int beta){
     int fw = ;
     for(int i=hsh[alpha];i;i=Nxt[i]){
     if(real[i] != beta) continue;
     fw += val[i];
     }
     return fw;
 }

 void work(){
     for(int i=;i<=n;i++) nouse[a[i]]++;
     for(int i=;i<=m;i++){
     if(Q[i].cas !=  || Q[i].l != ) continue;
     int fw = ; for(int j=;j<Q[i].str.length();j++){fw = (1ll*fw*nouse[Q[i].str[j]-''])%mod;}
     ans[Q[i].r] = fw;
     }
     memset(nouse,,sizeof(nouse)); nouse[] = ;nuse[] = ;
     for(int i=;i<=;i++) nouse[i] = (nouse[i-]*10ll)%hmod1,nuse[i] = (nuse[i-]*10ll)%hmod2;

     for(int i=;i<=m;i++){
     if(Q[i].cas == ) link(Q[i].l,Q[i].r);
     else if(Q[i].cas == ) cut(Q[i].l);
     else{
         if(Q[i].l == ) continue;
         int fw = ;long long hres1=,hres2 = ;
         for(int j=;j<Q[i].l;j++){
         hres1 = hres1*+Q[i].str[j]-'';
         hres2 = hres2*+Q[i].str[j]-'';
         hres1 %= hmod1; hres2 %= hmod2;
         }
         fw = (1ll*fw*COUNT(hres1,hres2))%mod;
         for(int j=Q[i].l;j<Q[i].str.length();j++){
         hres1 -= ((Q[i].str[j-Q[i].l]-'')*nouse[Q[i].l-])%hmod1;
         hres1 += hmod1; hres1 %= hmod1;
         hres2 -= ((Q[i].str[j-Q[i].l]-'')*nuse[Q[i].l-])%hmod2;
         hres2 += hmod2; hres2 %= hmod2;
         hres1 = (hres1*+Q[i].str[j]-'');hres1 %= hmod1;
         hres2 = (hres2*+Q[i].str[j]-'');hres2 %= hmod2;
         fw = (1ll*fw*COUNT(hres1,hres2))%mod;
         }
         ans[Q[i].r] = fw;
     }
     }
     for(int i=;i<=Num;i++) printf("%d\n",ans[i]);
 }

 int main(){
     read();
     work();
     return ;
 }