解题：APIO 2014 回文串

题面

初见SAM

洛谷数据太弱了，我SAM写错了居然有90pts=。=？？？

SAM求一个子串$(l,r)$的出现次数：从右端点对应状态开始在parent树上倍增，当目标节点的$len$大于等于子串长度时就往上跳，最后所在节点的$len$就是该串的出现次数

于是边$manacher$边在SAM上统计当前串的出现次数即可，复杂度$O(n\log n)$，注意边界

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=,K=;
 int p[N],noww[N],goal[N];
 int rnk[N],bkt[N],mul[N][K];
 int trs[N][],fth[N],len[N],siz[N],num[N];
 char str[N>>],Str[N];
 int lth,lst,cnt,tot,mid,maxr;
 long long ans;
 void link(int f,int t)
 {
     noww[++cnt]=p[f];
     goal[cnt]=t,p[f]=cnt;
 }
 void DFS(int nde,int fth)
 {
     mul[nde][]=fth;
     for(int i=p[nde];i;i=noww[i])
         DFS(goal[i],nde),siz[nde]+=siz[goal[i]];
     for(int i=;mul[nde][i];i++)
         mul[nde][i]=mul[mul[nde][i-]][i-];
 }
 void Insert(int ch,int ps)
 {
     int nde=lst,newn=++tot;
     num[ps]=lst=newn,siz[newn]=,len[newn]=len[nde]+;
     while(nde&&!trs[nde][ch])
         trs[nde][ch]=newn,nde=fth[nde];
     if(!nde) fth[newn]=;
     else
     {
         int tran=trs[nde][ch];
         if(len[tran]==len[nde]+)
             fth[newn]=tran;
         else
         {
             int rnde=++tot; len[rnde]=len[nde]+;
             for(int i=;i<=;i++) trs[rnde][i]=trs[tran][i];
             fth[rnde]=fth[tran],fth[tran]=fth[newn]=rnde;
             while(nde&&trs[nde][ch]==tran)
                 trs[nde][ch]=rnde,nde=fth[nde];
         }
     }
 }
 void prework()
 {
     register int i,j;
     scanf("%s",str+),lth=strlen(str+),lst=tot=;
     for(i=;i<=lth;i++) Insert(str[i]-'a',i);
     for(i=;i<=tot;i++) link(fth[i],i); DFS(,);
 }
 void Solve(int l,int r)
 {
     l=(l+l%)/,r=(r-r%)/;
     if(l>r) return ;
     int nde=num[r],lth=r-l+;
     for(int i=;~i;i--)
         if(lth<=len[mul[nde][i]])
             nde=mul[nde][i];
     ans=max(ans,1ll*lth*siz[nde]);
 }
 void Manacher()
 {
     register int i;
     int newl=*lth+;
     for(i=;i<=newl;i++)
         Str[i]=(i&)?'?':str[i>>];
     Str[]='>',Str[newl+]='<';
     for(i=;i<=newl;i++)
     {
         fth[i]=(maxr>=i)?min(maxr-i+,fth[*mid-i]):;
         Solve(i-fth[i]+,i+fth[i]-);
         while(Str[i-fth[i]]==Str[i+fth[i]])
             fth[i]++,Solve(i-fth[i]+,i+fth[i]-);
         if(i+fth[i]->maxr) maxr=i+fth[i]-,mid=i;
     }
 }
 int main()
 {
     prework(),Manacher();
     printf("%lld",ans);
     return ;
 }