树（prufer序，dp，排列组合）

Description

有nn个点，第ii个点的限制为度数不能超过aiai。现在对于每一个s(1≤s≤n)s(1≤s≤n)，问从这nn个点中选出ss个点组成有标号无根树的方案数mod1004535809mod1004535809。

Solution

参考博客

由于每一个prufer序对应了一棵唯一的树且prufer序中每个数的出现次数为该节点的度−1−1，考虑在prufer序上进行dp。

设dpi,j,kdpi,j,k表示前ii个节点中，选了jj个节点，一共填了kk位。

已知选了ss个节点，每个点的出现次数为CiCi，则方案数为：s!C1!C2!…Cs!s!C1!C2!…Cs!

为了方便，我们将dpi,j,kdpi,j,k除以s!s!，那么得到转移方程就非常显然了：

(f[i + 1][j][k] += f[i][j][k]) %= Ha;//不选第i+1个点
(f[i + 1][j + 1][k + l] += f[i][j][k] * ifac[l] % Ha) %= Ha;//将l个(i+1)点加入prufer序
//ifac[l]为l阶乘的逆元

Code

/**************************
Happy New Year!
Au: Hany01
Date: Jan 1st, 2018
Prob: tree
Email: hany01@foxmail.com
**************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir first
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Ha (1004535809)

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
register int _, __; register char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}

inline void File()
{
freopen("tree.in", "r", stdin);
freopen("tree.out", "w", stdout);
}

template<typename T> inline void mod(T &x) { if (x >= Ha) x -= Ha; }

const int maxn = 105;

int n, a[maxn];
ll fac[maxn], ifac[maxn], f[maxn][maxn][maxn];

inline ll Pow(ll a, ll b) { ll Ans = 1; for ( ; b; b >>= 1, a = a * a % Ha) if (b & 1) (Ans *= a) %= Ha; return Ans; }

inline void Init()
{
n = read();
For(i, 1, n) a[i] = read();
fac[0] = 1;
For(i, 1, n) fac[i] = fac[i - 1] * i % Ha;
ifac
= Pow(fac
, Ha - 2);
Fordown(i, n, 1) ifac[i - 1] = ifac[i] * i % Ha;
}

int main()
{
File();
Init();
//dp[i][j][k] means till the i_th node, you have choosed j kinds of them and k nodes altogether.
f[0][0][0] = 1;
rep(i, n) rep(j, i + 1) rep(k, n - 1) {
if (!f[i][j][k]) continue;
mod(f[i + 1][j][k] += f[i][j][k]);
rep(l, min(n - k - 1, a[i + 1]))
mod(f[i + 1][j + 1][k + l] += f[i][j][k] * ifac[l] % Ha);
}
printf("%d", n);
For(i, 2, n) cout << ' ' << f
[i][i - 2] * fac[i - 2] % Ha;
putchar('\n');
return 0;
}
//十年磨一剑，霜刃未曾试。
//    -- 贾岛《剑客 / 述剑》