【HDU4622】Reincarnation

【HDU4622】Reincarnation

一眼似乎不可做，但发现(strlen(x))很小，暴力(O(n^2))预处理每个区间((l,r))，查询时(O(1))输出就好了

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
typedef int LL;
const LL maxn=2010;
inline LL Read(){
	LL x=0,f=1; char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-') f=-1; c=getchar();
	}
	while(c>='0'&&c<='9')
	    x=(x<<3)+(x<<1)+c-'0',c=getchar();
	return x*f;
}
struct node{
	LL len,fail;
	LL son[26];
}t[maxn<<1];
LL n,m,T,nod,last;
LL ans[maxn][maxn];
char s[maxn];
inline void Insert(LL c){
	LL np=++nod,p=last;
	last=np;
	t[np].len=t[p].len+1;
	while(p&&!t[p].son[c]){
		t[p].son[c]=np,
		p=t[p].fail;
	}
	if(!p)
	    t[np].fail=1;
	else{
		LL q=t[p].son[c];
		if(t[q].len==t[p].len+1)
		    t[np].fail=q;
		else{
			LL nq=++nod;
		    t[nq]=t[q];
		    t[nq].len=t[p].len+1;
		    t[np].fail=t[q].fail=nq;
		    while(p&&t[p].son[c]==q){
		    	t[p].son[c]=nq,
		    	p=t[p].fail;
			}
		}
	}
}
int main(){
	T=Read();
	while(T--){
		scanf(" %s",s+1);
		LL Len=strlen(s+1);
		for(LL i=1;i<=Len;++i){
			nod=last=1;
			memset(t,0,sizeof(t));
			for(LL j=i;j<=Len;++j){
				Insert(s[j]-'a'),
				ans[i][j]=ans[i][j-1]+t[last].len-t[t[last].fail].len;
			}
		}
		m=Read();
		while(m--){
			LL l=Read(),r=Read();
			printf("%d
",ans[l][r]);
		}
	}
	return 0;
}