[校内训练2021_03_25]BC

题目大意：求一个 有标号树的 本质不同的 有根点分树 数量。要求平方。

思考：咕咕咕

#include<bits/stdc++.h>
#define mod 1000000007
using namespace std;
typedef long long int ll;
const int maxn=5E3+5;
int n;
int size,head[maxn];
ll f[maxn][maxn],C[maxn][maxn];
struct edge
{
    int to,next;
}E[maxn*2];
inline void add(int u,int v)
{
    E[++size].to=v;
    E[size].next=head[u];
    head[u]=size;
}
ll tmp[maxn];
int sum[maxn];
void dfs(int u,int F)
{
    f[u][0]=1;
    for(int e=head[u];e;e=E[e].next)
    {
        int v=E[e].to;
        if(v==F)
            continue;
        dfs(v,u);
        for(int i=0;i<=sum[u]+sum[v];++i)
            tmp[i]=0;
        for(int i=0;i<=sum[u];++i)
            for(int j=0;j<=sum[v];++j)
                tmp[i+j]=(tmp[i+j]+f[u][i]*f[v][j]%mod*C[i+j][i])%mod;
        sum[u]+=sum[v];
        for(int i=0;i<=sum[u];++i)
            f[u][i]=tmp[i];
    }
    memset(tmp,0,sizeof(tmp));
    ++sum[u];
    for(int i=1;i<=sum[u];++i)
        tmp[i]=f[u][i-1];
    for(int i=sum[u];i>=0;--i)
        f[u][i]=(f[u][i+1]+tmp[i])%mod;
}
int main()
{
    freopen("game.in","r",stdin);
    freopen("game.out","w",stdout);
    ios::sync_with_stdio(false);
    cin>>n;
    for(int i=2;i<=n;++i)
    {
        int x,y;
        cin>>x>>y;
        assert(x!=y);
        add(x,y);
        add(y,x);
    }
    for(int i=0;i<=n;++i)
    {
        C[i][0]=1;
        for(int j=1;j<=i;++j)
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
    }
    dfs(1,0);
    cout<<f[1][0]<<endl;
    return 0;
}

---

题目链接：https://www.luogu.com.cn/problem/P1393

题解：https://mivik.gitee.io/2020/solution/mivik-string-contest-title/

#include<bits/stdc++.h>
#define mod 998244353
#define G 3
#define Gi 332748118
using namespace std;
typedef long long int ll;
const int maxn=1E5+5;
ll n,m,k;
int a[maxn];
inline ll qpow(ll x,ll y)
{
    ll ans=1,base=x;
    while(y)
    {
        if(y&1)
            ans=ans*base%mod;
        base=base*base%mod;
        y>>=1;
    }
    return ans;
}
int fail[maxn];
namespace POLY
{
    vector<ll>pre[20],prei[20];
    inline void init()
    {
        for(int i=1;i<20;++i)
        {
            ll d=qpow(G,(mod-1)/(1<<i));
            for(ll j=0,w=1;j<(1<<i);++j,w=w*d%mod)
                pre[i].push_back(w);
            d=qpow(Gi,(mod-1)/(1<<i));
            for(ll j=0,w=1;j<(1<<i);++j,w=w*d%mod)
                prei[i].push_back(w);
        }
    }
    int r[(1<<20)+5];
    inline void DFT(ll*A,int limit)
    {
        for(int i=1;i<limit;++i)
            if(i<r[i])
                swap(A[i],A[r[i]]);
        for(int len=2,now=1;len<=limit;len<<=1,++now)
            for(int i=0;i<limit;i+=len)
                for(int j=0,p1=i,p2=i+len/2;j<len/2;++j,++p1,++p2)
                {
                    ll a=A[p1],b=A[p2]*pre[now][j]%mod;
                    A[p1]=a+b>=mod?a+b-mod:a+b;
                    A[p2]=a-b<0?a-b+mod:a-b;
                }
    }
    inline void IDFT(ll*A,int limit)
    {
        for(int i=1;i<limit;++i)
            if(i<r[i])
                swap(A[i],A[r[i]]);
        for(int len=2,now=1;len<=limit;len<<=1,++now)
            for(int i=0;i<limit;i+=len)
                for(int j=0,p1=i,p2=i+len/2;j<len/2;++j,++p1,++p2)
                {
                    ll a=A[p1],b=A[p2]*prei[now][j]%mod;
                    A[p1]=a+b>=mod?a+b-mod:a+b;
                    A[p2]=a-b<0?a-b+mod:a-b;
                }
    }
    ll tmp1[(1<<20)+5],tmp2[(1<<20)+5];
    inline void FFT(ll*A,ll*B,int n1,int n2,ll*to)
    {
        int limit=1;
        while(limit<n1+n2+1)
            limit<<=1;
        for(int i=0;i<limit;++i)
            tmp1[i]=tmp2[i]=0;
        for(int i=0;i<=n1;++i)
            tmp1[i]=A[i];
        for(int i=0;i<=n2;++i)
            tmp2[i]=B[i];
        for(int i=0;i<limit;++i)
            r[i]=(r[i>>1]>>1)|((i&1)?(limit>>1):0);
        DFT(tmp1,limit);
        DFT(tmp2,limit);
        for(int i=0;i<limit;++i)
            tmp1[i]=tmp1[i]*tmp2[i]%mod;
        IDFT(tmp1,limit);
        ll base=qpow(limit,mod-2);
        for(int i=0;i<=n1+n2;++i)
            to[i]=tmp1[i]*base%mod;
    }
    ll tmp3[(1<<20)+5],tmp4[(1<<20)+5],tmp5[(1<<20)+5];
    inline void inv(ll*A,int n)
    {
        tmp3[0]=qpow(A[0],mod-2);
        for(int now=1;now<n+1;now<<=1)
        {
            int limit=now<<2;
            for(int i=0;i<limit;++i)
                tmp4[i]=tmp5[i]=0;
            for(int i=0;i<now;++i)
                tmp4[i]=tmp3[i];
            for(int i=0;i<(now<<1);++i)
                tmp5[i]=A[i];
            for(int i=0;i<limit;++i)
                r[i]=(r[i>>1]>>1)|((i&1)?(limit>>1):0);
            DFT(tmp4,limit);
            DFT(tmp5,limit);
            for(int i=0;i<limit;++i)
                tmp3[i]=((-tmp4[i]*tmp4[i]%mod*tmp5[i]%mod+tmp4[i]*2)%mod+mod)%mod;
            IDFT(tmp3,limit);
            ll base=qpow(limit,mod-2);
            for(int i=0;i<(now<<1);++i)
                tmp3[i]=tmp3[i]*base%mod;
        }
        for(int i=0;i<=n;++i)
            A[i]=tmp3[i];
    }
}
ll C[maxn];
inline void solve()
{
    int pos=0;
    for(int i=2;i<=k;++i)
    {
        while(pos&&a[i]!=a[pos+1])
            pos=fail[pos];
        if(a[i]==a[pos+1])
            ++pos;
        fail[i]=pos;
    }
    for(int p=k;p;p=fail[p])
        C[k-p]=1;
    for(int i=n;i>=1;--i)
        C[i]=(C[i]-C[i-1]*m%mod+mod)%mod;
    ++C[k];
    POLY::inv(C,n);
    for(int i=n;i>=k;--i)
        C[i]=C[i-k];
    ll ans=0;
    for(int i=k;i<=n;++i)
        ans=(ans+C[i]*qpow(m,n-i))%mod;
    ans=(ans%mod+mod)%mod;
    ans=ans*qpow(qpow(m,n),mod-2)%mod;
    cout<<ans<<endl;
}
inline int read()
{
    char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    int s=ch-'0';ch=getchar();
    while(isdigit(ch)){s=s*10+ch-'0';ch=getchar();}
    return s;
}
int main()
{
    freopen("bye.in","r",stdin);
    freopen("bye.out","w",stdout);
    ios::sync_with_stdio(false);
    POLY::init();
    n=read(),m=read(),k=read();
    for(int i=1;i<=k;++i)
        a[i]=read();
    solve();
    return 0;
}