BZOJ5305 HAOI2018苹果树（概率期望+动态规划）

　　每种父亲编号小于儿子编号的有标号二叉树的出现概率是相同的，问题相当于求所有n个点的此种树的所有结点两两距离之和。

　　设f[n]为答案，g[n]为所有此种树所有结点的深度之和，h[n]为此种树的个数。

　　枚举左右子树大小，则有f[n]=Σ{[f[i]+(g[i]+h[i]*i)·(n-i)]·h[n-i-1]+[f[n-i-1]+(g[n-i-1]+h[n-i-1]*(n-i-1))·(i+1)]·h[i]}·C(n-1,i)，即对两棵子树分别统计贡献，C(n-1,i)即给左右子树分配编号。g[n]=Σ[(g[i]+h[i]*i)·h[n-i-1]+(g[n-i-1]+h[n-i-1]*(n-i-1))·h[i]]·C(n-1,i)，h[n]=Σh[i]·h[n-i-1]·C(n-1,i)，比较显然。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 2010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,P,C[N][N],f[N],g[N],h[N];
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5305.in","r",stdin);
    freopen("bzoj5305.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read(),P=read();
    C[0][0]=1;
    for (int i=1;i<=n;i++)
    {
        C[i][0]=C[i][i]=1;
        for (int j=1;j<i;j++) 
        C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;
    }
    f[0]=g[0]=0;h[0]=1;
    for (int i=1;i<=n;i++)
    {
        for (int j=0;j<i;j++)
        inc(h[i],1ll*h[j]*h[i-j-1]%P*C[i-1][j]%P);
        for (int j=0;j<i;j++)
        inc(g[i],((g[j]+1ll*j*h[j])%P*h[i-j-1]+(g[i-j-1]+1ll*(i-j-1)*h[i-j-1])%P*h[j])%P*C[i-1][j]%P);
        for (int j=0;j<i;j++)
        inc(f[i],((f[j]+(g[j]+1ll*j*h[j])%P*(i-j))%P*h[i-j-1]+(f[i-j-1]+(g[i-j-1]+1ll*(i-j-1)*h[i-j-1])%P*(j+1))%P*h[j])%P*C[i-1][j]%P);
    }
    cout<<f[n];
    return 0;
}