不过求
g
g
g不用
O
(
n
2
)
D
P
O(n^2)DP
O(n2)DP，
g
[
n
]
g[n]
g[n]直接就是卡特兰数的第
n
−
1
n-1
n−1项。即：
g
[
n
]
=
(
2
(
n
−
1
)
n
−
1
)
−
(
2
(
n
−
1
)
n
−
2
)
g[n]=\binom{2(n-1)}{n-1}-\binom{2(n-1)}{n-2}
g[n]=(n−12(n−1))−(n−22(n−1))
相当于在平面直角坐标系中，要从
(
0
,
0
)
(0,0)
(0,0)走到
(
n
,
n
)
(n,n)
(n,n)，有一条线段
y
=
x
(
x
∈
(
0
,
n
)
)
y=x(x\in(0,n))
y=x(x∈(0,n))不能触碰，注意是开区间。所以卡特兰数/组合数的计算方法就行了。
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 105;
int n, mod, g[MAXN], f[MAXN][MAXN][MAXN], fac[MAXN<<1], inv[MAXN<<1];
int C(int n, int m) { return m > n ? 0 : 1ll * fac[n] * inv[m] % mod * inv[n-m] % mod; }
int main(){
scanf("%d%d", &n, &mod);
fac[0] = inv[0] = inv[1] = fac[1] = 1;
for(int i = 2; i <= (n<<1); ++i){
fac[i] = 1ll * fac[i-1] * i %mod;
inv[i] = 1ll * (mod - mod/i) * inv[mod%i] % mod;
}
for(int i = 2; i <= (n<<1); ++i) inv[i] = 1ll * inv[i-1] * inv[i] % mod;
for(int i = 1; i <= n; ++i) g[i] = C(2*i - 2, i - 1) - C(2*i - 2, i - 2);
f[0][0][0] = 1;
for(int i = 0; i <= n; ++i)
for(int j = 0; j <= n; ++j) if(i || j) {
for(int k = max(0, i+j-n); k <= i && k <= j; ++k){
int &ret = f[i][j][k];
if(i && k) ret = (ret + 1ll * f[i-1][j][k-1] * (j-k+1)) % mod;
if(j && k) ret = (ret + 1ll * f[i][j-1][k-1] * (i-k+1)) % mod;
if(i) ret = (ret + 1ll * f[i-1][j][k] * (n - (i-1+j-k))) % mod;
if(j) ret = (ret + 1ll * f[i][j-1][k] * (n - (i+j-1-k))) % mod;
for(int d = 1; d <= k; ++d)
ret = (ret - 1ll * f[i-d][j-d][k-d] * C(n-(i+j-k-d), d) % mod * g[d] % mod * fac[d] % mod) % mod;
}
}
printf("%d\n", (f[n][n][n] + mod) % mod);
}
没做过这种类型的感觉好难。。
