后缀树（BZOJ3238TLE）

#include<cstdio>
#include<cstring>
#define LL long long 

 char st[500010],stt[500010];
 LL ans;
 int leafr,ndcnt=1,sidcnt,lastcre;
 int isleaf[1000010],dep[1000010],nd[1000010],l[1000010],r[1000010],des[1000010],
       next[1000010],sufchain[1000010];

  
 int getr(int sid){
     if (isleaf[des[sid]]) return(leafr);else return(r[sid]);
 }
 
 void cut(int po,int sid,int len){
     dep[++ndcnt]=dep[po]+len;
     isleaf[ndcnt]=0;
     
     nd[ndcnt]=++sidcnt;
     des[sidcnt]=des[sid];
     l[sidcnt]=len+l[sid];r[sidcnt]=getr(sid);next[sidcnt]=-1;des[sidcnt]=des[sid];
    des[sid]=ndcnt;r[sid]=len+l[sid]-1;
 }
 
 void ins(int po,int alph){
     dep[++ndcnt]=dep[po]+1;
     isleaf[ndcnt]=1;
     nd[ndcnt]=-1;
     
     l[++sidcnt]=alph;r[sidcnt]=alph;
     des[sidcnt]=ndcnt;next[sidcnt]=nd[po];nd[po]=sidcnt;
 }

 void buildtree(){
    int po=1,lastins=1;nd[1]=-1;sufchain[1]=1;int prom=0;
    
    for (int i=1;i<=strlen(st)-1;i++){
        leafr=i;int p,flag;
        prom=0;
        
        while (lastins<=i){
            
            while(1){
                flag=0;
                for (p=nd[po];p!=-1;p=next[p])
                if (st[l[p]]==st[lastins+dep[po]]) break;
                
                if (p==-1){ins(po,i);
                           if (prom) sufchain[lastcre]=po;
                           prom=0;
                           lastcre=po;
                           break;}
                if (i-lastins+1-dep[po]<=getr(p)-l[p]+1){
                    if (st[i]==st[l[p]+i-lastins-dep[po]]) 
                      {flag=1;
                      if (prom) sufchain[lastcre]=po;
                      prom=0;
                      lastcre=po;
                      break;}
                    if (st[i]!=st[l[p]+i-lastins-dep[po]]) {
                        cut(po,p,i-lastins-dep[po]);
                        if (prom) sufchain[lastcre]=ndcnt;
                        if (i-lastins-dep[po]>1) prom=1;else prom=0;
                        lastcre=ndcnt;sufchain[lastcre]=1;
                        ins(ndcnt,i);break;
                    }
                }
                if (i-lastins+1-dep[po]>getr(p)-l[p]+1) po=des[p];
            }
            if (flag) {break;}
            po=sufchain[po];lastins++;
      }
    }
 }
 
 int top=0;
 struct str{
     int po,fr;
     LL cnt,len; 
 }sta[500010];
 
      
     void pb(int ps,int le,int p){
         sta[++top].po=p;sta[top].len=le;sta[top].fr=ps;
         sta[top].cnt=0;
     }
     
    void pop(){
        if (isleaf[sta[top].po]) sta[top].cnt++;
        sta[sta[top].fr].cnt+=sta[top].cnt;
        top--;
    }
 
 void getans(){
     pb(0,0,1);top=1;
     
     int i=1;
     while (top>=i){
         int tp=i;
         for (int p=nd[sta[tp].po];p!=-1;p=next[p]) 
         pb(tp,getr(p)-l[p]+1,des[p]);
         i++;
     }
     
    while(top){
        ans-=sta[top].len*sta[top].cnt*(sta[top].cnt-1);
        pop();
    } 
 }

 int main(){
     scanf("%s",stt);
     
     LL len=strlen(stt);ans=(len-1)*len*(len+1)/2;
    
    st[0]=' ';
    strcat(stt,"$");strcat(st,stt);
    
    buildtree();

    getans();
    
    printf("%lld",ans);
  }

注意某一个串对应点到后缀树根的深度是$O(n \cdot \sqrt{n})$的