求不同子串個數

裸的後綴自動機

 1 #include<cstring>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cstdio>
 6 
 7 #define ll long long
 8 #define N 2000007
 9 using namespace std;
10 inline int read()
11 {
12     int x=0,f=1;char ch=getchar();
 
13     while(!isdigit(ch)){if(ch==‘-‘)f=-1;ch=getchar();}
14     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();}
15     return x*f;
16 }
17 
18 int n;
19 ll ans;
20 struct sam
21 {
22     int last,cnt;
23     int c[N][26],fa[N],mx[N];
24     sam(){last=cnt=1;}
25     void
 
 extend(int x)
26     {
27         int p=last,np=last=++cnt;mx[np]=mx[p]+1;
28         while(p&&!c[p][x])
29         {
30             c[p][x]=np;
31             p=fa[p];
32         }
33         if(!p)fa[np]=1;
34         else
35         {
36             int q=c[p][x];
37             if
 
(mx[q]==mx[p]+1)fa[np]=q;
38             else
39             {
40                 int nq=++cnt;mx[nq]=mx[p]+1;
41                 memcpy(c[nq],c[q],sizeof(c[q]));
42                 fa[nq]=fa[q];
43                 fa[q]=fa[np]=nq;
44                 while(c[p][x]==q)c[p][x]=nq,p=fa[p];
45             }
46         }
47     }
48 }sam;
49 char s[N];
50 
51 int main()
52 {
53     scanf("%s",s+1);n=strlen(s+1);
54     for (int i=1;i<=n;i++)
55         sam.extend(s[i]-‘a‘);
56     for (int i=1;i<=sam.cnt;i++)
57         ans+=sam.mx[i]-sam.mx[sam.fa[i]];
58     printf("%lld",ans);
59 }

hihocoder 後綴自動機二·重復旋律5