题意
给你一个串,求不同字串个数。
(nle10^5)
sol
直接建SAM然后输出(sum_{i=1}^{tot}len[i]-len[fa[i]])
code
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 1e5+5;
int T,n,last=1,tot=1,tr[N][26],fa[N],len[N];
char s[N];long long ans;
void extend(int c)
{
int v=last,u=++tot;last=u;
len[u]=len[v]+1;
while (v&&!tr[v][c]) tr[v][c]=u,v=fa[v];
if (!v) fa[u]=1;
else
{
int x=tr[v][c];
if (len[x]==len[v]+1) fa[u]=x;
else
{
int y=++tot;
memcpy(tr[y],tr[x],sizeof(tr[y]));
fa[y]=fa[x];fa[x]=fa[u]=y;len[y]=len[v]+1;
while (v&&tr[v][c]==x) tr[v][c]=y,v=fa[v];
}
}
}
int main()
{
scanf("%d",&T);
while (T--)
{
scanf("%s",s+1);n=strlen(s+1);
ans=0;last=tot=1;
memset(tr,0,sizeof(tr));
memset(fa,0,sizeof(fa));
memset(len,0,sizeof(len));
for (int i=1;i<=n;++i) extend(s[i]-'A');
for (int i=1;i<=tot;++i) ans+=len[i]-len[fa[i]];
printf("%lld
",ans);
}
return 0;
}