958F3 - Lightsabers (hard)

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从$n$种有颜色的球里面取$k$，相同颜色算一种，一个颜色多球算一种。有$m$种颜色，$n,m\leq 2\times 10^5$

假设某种颜色有$y$个，它的生成函数$=1+x+x^2..x^y$

启发式合并$FFT$，即先将长度小的多项式项乘放到堆里面。

  • 所得出的多项式保持长度$\leq k$
  • 根默认的优先级就是大根堆
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 6e5 + 10;
const int mod = 1009;
const double pi = acos(-1.0);

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 - '0', c = getchar();
    return x * f;
}

struct Complex
{
    double x, y;
    Complex(double _x = 0.0, double _y = 0.0)
    {
        x = _x;
        y = _y;
    }

    Complex operator-(const Complex &b) const
    {
        return Complex(x - b.x, y - b.y);
    }

    Complex operator+(const Complex &b) const
    {
        return Complex(x + b.x, y + b.y);
    }

    Complex operator*(const Complex &b) const
    {
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
};
int rev[N];
void FFT(Complex *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        Complex temp(cos(pi / l), inv * sin(pi / l));
        for (int i = 0; i < n; i += (l << 1))
        {
            Complex omega(1, 0);
            for (int j = 0; j < l; j++, omega = omega * temp)
            {
                Complex x = A[i + j], y = omega * A[i + j + l];
                A[i + j] = x + y;
                A[i + j + l] = x - y;
            }
        }
    }

    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i].x = ll(A[i].x / n + 0.5) % mod;
}
int Ployinit(int n, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    return ML;
}
void FFTX(Complex *a, int n, Complex *b, int m)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    FFT(a, ML, 1);
    FFT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        a[i] = a[i] * b[i];
    FFT(a, ML, -1);
}
struct Ploy
{
    int id;
    int siz;
    bool operator<(const Ploy &X) const
    {
        return siz > X.siz;
    }
};
vector<int> h[N];
priority_queue<Ploy> q;
vector<int> p;
Complex a[N], b[N];
int main()
{
    int n = read(), m = read(), k = read();
    for (int i = 1; i <= m; i++)
        h[i].push_back(1);
    for (int i = 1; i <= n; i++)
    {
        int x = read();
        h[x].push_back(1);
    }
    for (int i = 1; i <= m; i++)
    {
        while (h[i].size() > k + 1)
            h[i].pop_back();
        if (h[i].size() > 1)
            q.push((Ploy){i, h[i].size()});
    }
    int cnt = m;
    while (q.size() >= 2)
    {
        Ploy f = q.top();
        q.pop();
        Ploy g = q.top();
        q.pop();
        int len1 = f.siz;
        int len2 = g.siz;
        int ML = Ployinit(len1, len2);
        for (int i = 0; i < ML; i++)
            a[i].x = a[i].y = b[i].x = b[i].y = 0;
        for (int i = 0; i < len1; i++)
            a[i].x = h[f.id][i];
        for (int i = 0; i < len2; i++)
            b[i].x = h[g.id][i];
        //cout << n << " " << m << endl;
        FFTX(a, len1, b, len2);
        ++cnt;
        for (int i = 0; i <= min(len1 + len2, k + 2); i++)
            h[cnt].push_back(a[i].x);
        q.push((Ploy){cnt, h[cnt].size()});
    }
    Ploy f = q.top();
    q.pop();
    printf("%d\n", h[f.id][k]);
}