【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;
}