Codeforces Round #542 [Alex Lopashev Thanks-Round] (Div. 1)C. Morse Code

题意:给你n个01字符,每次问你前缀的所有本质不同的子串,由摩斯密码组成的方案数和.

题解:离线处理,把字符建sam,通过topo序来dp计算每个节点表示的子串方案数的和.统计答案时,把n个字符挨个匹配过去,把所有经过的节点加上方案数,还有每个节点的fa链,也要统计(因为是当前串的后缀),用一个vis标记访问过没有.复杂度O(n*32),于是这题n能出到1e6

//#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-7;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=3000+10,maxn=100000+10,inf=0x3f3f3f3f;

int n,b[N];
char s[N];
struct SAM{
    int last,cnt;
    int ch[N<<1][2],fa[N<<1],l[N<<1];
    int a[N<<1],c[N<<1];
    bool vis[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;
            }
        }
    }
    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;
        dp[1]=1;
    }
    void dfs(int u,int dep,ll x,int now)
    {
        if(dep==0&&(now==3||now==5||now==14||now==15));
        else if(dep!=4)add(dp[u],x);
        if(dep==0)return ;
        for(int i=0;i<2;i++)if(ch[u][i])
            dfs(ch[u][i],dep-1,x,now*2+i);
    }
    void cal()
    {
        topo();
        for(int i=1;i<=cnt;i++)dfs(a[i],4,dp[a[i]],0);
        int now=1;ll ans=0;
        vis[1]=1;
        for(int i=1;i<=n;i++)
        {
            now=ch[now][b[i]];
            add(ans,dp[now]);vis[now]=1;
            int te=fa[now];
            while(te&&!vis[te])
            {
                vis[te]=1,add(ans,dp[te]);
                te=fa[te];
            }
            printf("%lld\n",ans);
        }
    }
}sam;
int main()
{
    scanf("%d",&n);
    sam.build();
    for(int i=1,x;i<=n;i++)scanf("%d",&b[i]),sam.ins(b[i],i);
    sam.cal();
    return 0;
}
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