LuoGu: CF995F Cowmpany Cowmpensation
CF: F. Cowmpany Cowmpensation
考虑求出恰好使用了 $i$ 个值的方案数 $g _ i$ ，那么有答案 $\sum _ {i = 1} ^ n \binom m i g _ i$ ，其中 $\binom m i$ 可以 $O(n)$ 暴力算。
注意到虽然 $m$ 很大，但是实际上用到的不超过 $n$ 个，那么考虑 DP ，$f _ {i, j}$ 表示在 $i$ 点，选择了第 $j$ 大的数的方案数。方程 $f _ {u, i} = \prod _ v \sum _ {k = 1} ^ i f _ {v, k}$ 。枚举哪些值可以和前一个值一样，那么可以发现 $\sum _ {k = 1} ^ i \binom {i - 1} {i - k} g _ i = f _ {1, i}$ 。递推得到答案。
查看代码
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| #include <cstdio>
#include <vector>
#define pb push_back
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 3e3 + 10, mod = 1e9 + 7;
vector <int> e[N];
int n, m, f[N][N], g[N];
int inv[N], fact[N], ifact[N];
int C (int a, int b) { return a < b ? 0 : (LL)fact[a] * ifact[b] % mod * ifact[a - b] % mod; }
void init ()
{
inv[1] = 1;
for (int i = 2; i < N; ++i)
inv[i] = -(LL)(mod / i) * inv[mod % i] % mod;
fact[0] = ifact[0] = 1;
for (int i = 1; i < N; ++i)
{
fact[i] = (LL)fact[i - 1] * i % mod;
ifact[i] = (LL)ifact[i - 1] * inv[i] % mod;
}
}
void dfs (int u)
{
for (int i = 1; i <= n; ++i) f[u][i] = 1;
for (int v : e[u])
{
dfs(v);
for (int k = 1; k <= n; ++k)
f[u][k] = (LL)f[u][k] * f[v][k] % mod;
}
for (int i = 2; i <= n; ++i)
f[u][i] = (f[u][i] + f[u][i - 1]) % mod;
}
int main ()
{
init ();
read(n, m);
for (int i = 2, a; i <= n; ++i)
read(a), e[a].pb(i);
dfs(1);
for (int i = 1; i <= n; ++i)
{
g[i] = (f[1][i] - f[1][i - 1]) % mod;
for (int j = 1; j < i; ++j)
g[i] = (g[i] - (LL)C(i - 1, i - j) * g[j]) % mod;
}
int res = 0;
for (int i = 1; i <= min(n, m); ++i)
{
int t = ifact[i];
for (int j = 0; j < i; ++j)
t = (LL)t * (m - j) % mod;
res = (res + (LL)t * g[i]) % mod;
}
write((res + mod) % mod);
return 0;
}
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