MemSQL Start[c]UP 2.0

题意:给3个字符串,问从1到min(l1,l2,l3)的长度的子串,找到从该位置长度为l,三个子串相同的三元组的个数
题解:把3个子串用分隔符串起来.然后分开统计每个节点在三个串中出现次数.最后乘起来就是该节点表示的三元组个数,然后l[fa[i]]+1到l[i]有贡献,对l差分一下就好了

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=300000+10,maxn=1000000+10,inf=0x3f3f3f3f;

char s[N];
struct SAM{
    int last,cnt;
    int ch[N<<1][27],fa[N<<1],l[N<<1],sz[N<<1][3];
    int a[N<<1],c[N<<1];
    ll dp[N<<1];
    void ins(int c,int id){
        int p=last,np=++cnt;last=np;l[np]=l[p]+1;
        for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
        if(!p)fa[np]=1;
        else
        {
            int q=ch[p][c];
            if(l[p]+1==l[q])fa[np]=q;
            else
            {
                int nq=++cnt;l[nq]=l[p]+1;
                memcpy(ch[nq],ch[q],sizeof(ch[q]));
                fa[nq]=fa[q];fa[q]=fa[np]=nq;
                for(;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
            }
        }
        if(id>=0)sz[np][id]=1;
    }
    void topo()
    {
        for(int i=1;i<=cnt;i++)c[l[i]]++;
        for(int i=1;i<=cnt;i++)c[i]+=c[i-1];
        for(int i=1;i<=cnt;i++)a[c[l[i]]--]=i;
    }
    void build(){
        last=cnt=1;
        int mi=inf;
        for(int i=0;i<3;i++)
        {
            scanf("%s",s);
            mi=min(mi,(int)strlen(s));
            for(int j=0;s[j];j++)ins(s[j]-'a',i);
            ins(26,-1);
        }
        topo();
        for(int i=cnt;i;i--)for(int j=0;j<3;j++)sz[fa[a[i]]][j]+=sz[a[i]][j];
        for(int i=1;i<=cnt;i++)
        {
            ll te=1ll*sz[i][0]*sz[i][1]%mod*sz[i][2]%mod;
            add(dp[l[fa[i]]+1],te);
            sub(dp[l[i]+1],te);
        }
        for(int i=1;i<=mi;i++)add(dp[i+1],dp[i]),printf("%lld
",dp[i]);
    }
}sam;
int main()
{
    sam.build();
    return 0;
}
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