[Ahoi2013]差异 bzoj 3238 后缀自动机

题目

分析

code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define N 500005
#define LL long long

using namespace std;

int n;
int len[N*2];
int ch[N*2][30],fail[N*2];
int f[N*2],w[N*2],b[N*2],c[N*2];
int last,root,cnt;
char s
;

void ins(int x)
{
int p,q,np,nq;
p=last; last=np=++cnt;
len[np]=len[p]+1; w[np]=f[np]=1;

for (;!ch[p][x]&&p;p=fail[p])
ch[p][x]=np;

if (p==0)
fail[np]=root;
else
{
q=ch[p][x];
if (len[q]==(len[p]+1))
fail[np]=q;
else
{
nq=++cnt;
len[nq]=len[p]+1;
for (int i=0;i<=26;i++) ch[nq][i]=ch[q][i];
fail[nq]=fail[q];
fail[q]=fail[np]=nq;
for (;ch[p][x]==q;p=fail[p])
ch[p][x]=nq;
}
}
}

LL solve()
{
for (int i=1;i<=cnt;i++) b[len[i]]++;
for (int i=1;i<=n;i++) b[i]+=b[i-1];
for (int i=1;i<=cnt;i++) c[b[len[i]]--]=i;
for (int i=cnt;i>=1;i--) f[fail[c[i]]]+=f[c[i]];
LL ans=0;
for (int i=1;i<=cnt;i++)
{
ans+=(LL)w[fail[i]]*f[i]*len[fail[i]];
w[fail[i]]+=f[i];
}
return ans;
}

int main()
{
scanf("%s",s+1);
n=strlen(s+1);
last=cnt=root=1;
for (int i=1;i<=n;i++) ins(s[i]-'a');
LL ans=0;
for (int i=1;i<=n;i++)
ans+=(LL)i*(n-1);
printf("%lld",ans-solve()*2);
return 0;
}