【SPOJ】Distinct Substrings

【SPOJ】Distinct Substrings

求不同子串数量

统计每个点有效的字符串数量（第一次出现的）

\(\sum\limits_{now=1}^{nod}now.longest-parents.longest\)

My complete code

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL maxn=3000;
LL nod,last,n,T;
LL len[maxn],fail[maxn],son[maxn][26];
char s[maxn];
inline void Insert(LL c){
    LL np=++nod,p=last;
    len[np]=len[p]+1;
    last=np;
    while(p&&!son[p][c]){
        son[p][c]=np,
        p=fail[p];
    }
    if(!p)
        fail[np]=1;
    else{
        LL q=son[p][c];
        if(len[q]==len[p]+1)
            fail[np]=q;
        else{
            LL nq=++nod;
            len[nq]=len[p]+1;
            fail[nq]=fail[q];
            memcpy(son[nq],son[q],sizeof(son[q]));
            fail[np]=fail[q]=nq;
            while(p&&son[p][c]==q){
                son[p][c]=nq,
                p=fail[p];
            }
        }
    }
}
int main(){
    scanf("%lld",&T);
    while(T--){
        memset(son,0,sizeof(son));
        nod=last=1;
        scanf(" %s",s);
        for(LL i=0;i<strlen(s);++i)
            Insert(s[i]-'A');
        LL ans=0;
        for(LL i=1;i<=nod;++i)
            ans+=len[i]-len[fail[i]];
        printf("%lld\n",ans);
    }
    return 0;
}