4566: [Haoi2016]找相同字符 广义后缀自动机

建立广义后缀自动机，维护两个串的$size$
那么答案等于$\sum (len_i-len_{fa_i})\times size_0 \times size_1$

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int N=200005;
char str1[N],str2[N];
int n1,n2,S,last,cnt;
long long ans;
int fa[N<<2],len[N<<2],ch[N<<2][26],size[2][N<<2],cc[N<<2],q[N<<2];

inline void extend(char c)
{
    int p=last,np=++cnt; last=np;
    len[np]=len[p]+1;
    while (p&&!ch[p][c]) ch[p][c]=np,p=fa[p];
    if (!p) fa[np]=S;
    else
    {
        int q=ch[p][c];
        if (len[p]+1==len[q]) fa[np]=q;
        else
        {
            int nq=++cnt;
            memcpy(ch[nq],ch[q],sizeof(ch[q]));
            len[nq]=len[p]+1;
            fa[nq]=fa[q];
            fa[q]=fa[np]=nq;
            while (ch[p][c]==q) ch[p][c]=nq,p=fa[p];
        }
    }
}

int main()
{
    scanf("%s",str1);
    scanf("%s",str2);
    n1=strlen(str1);
    n2=strlen(str2);
    S=last=++cnt;
    int p=S;
    for (int i=0;i<n1;i++)
    {
        extend(str1[i]-'a');
        p=ch[p][str1[i]-'a'];
        size[0][p]++;
    }
    last=S;
    p=S;
    for (int i=0;i<n2;i++)
    {
        extend(str2[i]-'a');
        p=ch[p][str2[i]-'a'];
        size[1][p]++;
    }
    for (int i=1;i<=cnt;i++) ++cc[len[i]];
    for (int i=1;i<=cnt;i++) cc[i]+=cc[i-1];
    for (int i=1;i<=cnt;i++) q[cc[len[i]]--]=i;
    for (int i=cnt;i;i--) size[0][fa[q[i]]]+=size[0][q[i]],size[1][fa[q[i]]]+=size[1][q[i]];
    for (int i=1;i<=cnt;i++)
        ans+=(long long)(len[i]-len[fa[i]])*size[0][i]*size[1][i];
    cout << ans << endl;
    return 0;
}