HDU 5552 (CDQ分治 NTT)

题意：n个点有环连通图计数。每条边可以染成m种颜色。

定义一些数组：
f[n]: n个点的连通图；
g[n]:n个点的图；
h[n]:n个点的树。
后两者计算很简单，每条边都有m+1种选择（不取这条边或者m种颜色），一共有n∗(n−1)2条边，所以g[n]=(m+1)n∗(n−1)2；n个点的树根据prufer数列一个树对应一个n-2长度的序列，而一个树有n-1条边，故h[n]=nn−2∗mn−1.

对于f数组，联通图的个数=图的总数−非联通图的个数，所以枚举1这个点所在的联通块的大小：

f[n]=g[n]−∑i=1n−1Ci−1n−1f[i]g[n−i]=g[n]−∑i=1n−1(n−1)!(i−1)!∗(n−i)!=g[n]−(n−1)!∑i=1n−1f[i](i−1)!∗g[n−i](n−i)!

这个东西是一个卷积的形式，用fft或者ntt加速就好了。

那么最后的答案就是n个点的连通图−n个点的树，也就是ans=f[n]−h[n]。

然后fft开心的跪了精度，所以用ntt来避免精度误差，原根的算出来是106。

FFT：（精度会炸）

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <map>#include <vector>#include <stack>using namespace std;#define mod 152076289#define pi acos(-1.0)#define maxn 40005struct plex {    long double x, y;    plex (long double _x = 0.0, long double _y = 0.0) : x (_x), y (_y) {}    plex operator + (const plex &a) const {        return plex (x+a.x, y+a.y);    }    plex operator - (const plex &a) const {        return plex (x-a.x, y-a.y);    }    plex operator * (const plex &a) const {        return plex (x*a.x-y*a.y, x*a.y+y*a.x);    }}x1[maxn], x2[maxn];void change (plex *y, int len) {    int i, j, k;    for(i = 1, j = len / 2; i < len - 1; i++) {        if (i < j) swap(y[i], y[j]);        k = len / 2;        while (j >= k) {            j -= k;            k /= 2;        }        if (j < k) j += k;    }}void fft(plex y[],int len,int on){    change(y,len);    for(int h = 2; h <= len; h <<= 1)    {        plex wn(cos(-on*2*pi/h),sin(-on*2*pi/h));        for(int j = 0;j < len;j+=h)        {            plex w(1,0);            for(int k = j;k < j+h/2;k++)            {                plex u = y[k];                plex t = w*y[k+h/2];                y[k] = u+t;                y[k+h/2] = u-t;                w = w*wn;            }        }    }    if(on == -1)        for(int i = 0;i < len;i++)            y[i].x /= len;}long long qpow (long long a, long long b) {      long long ret=1;      while (b) {          if (b&1) ret = (ret*a)%mod;          a = (a*a)%mod;          b >>= 1;      }      return ret;  }  long long n, m;long long f[maxn], g[maxn], h[maxn];//n个点的连通图 n个点的图 n个点的树long long c[maxn], fac[maxn], rev[maxn];void solve (int l, int r) {    if (l == r) {        f[l] += g[l];        f[l] %= mod;        return;    }    int mid = (l+r) >> 1;    solve (l, mid);    int len = 1;    while (len <= r-l+1) {len <<= 1;}    for (int i = 0; i < len; i++) {        x1[i] = x2[i] = plex (0, 0);    }    for (int i = l; i <= mid; i++) {        long long tmp = f[i]*rev[i-1]%mod;        x1[i-l] = plex (tmp, 0);    }    for (int i = 1; i <= r-l; i++) {        long long tmp = g[i]*rev[i]%mod;        x2[i-1] = plex (tmp, 0);    }    fft (x1, len, 1), fft (x2, len, 1);    for (int i = 0; i < len; i++) x1[i] = x1[i] * x2[i];    fft (x1, len, -1);    for (int i = mid+1; i <= r; i++) {        f[i] -= (long long) (x1[i-l-1].x + 0.5) %mod *fac[i-1] %mod;        (f[i] += mod) %= mod;    }    //cout << l << " " << r << ": ";    //for (int i = 1; i <= n; i++) cout << f[i] << " "; cout << endl;    solve (mid+1, r);}int main () {    int t, kase = 0;    fac[0] = 1;    for (int i = 1; i < maxn; i++) fac[i] = fac[i-1] * i % mod;    rev[maxn-1] = qpow (fac[maxn-1], mod - 2);    for (int i = maxn-2; i >= 0; i--) rev[i] = rev[i+1] * (i+1) % mod;    scanf ("%d", &t);    while (t--) {        scanf ("%lld%lld", &n, &m);        memset (f, 0, sizeof f);        c[0] = 1;        for (int i = 1; i < n; i++) {            c[i] = c[i-1]*(n-1-i+1)%mod*rev[i]%mod;        }        for (int i = 1; i <= n; i++) {            g[i] = qpow (1+m, i*(i-1)/2);        }        //for (int i = 1; i <= n; i++) cout << c[i] << " "; cout << endl;        //for (int i = 1; i <= n; i++) cout << g[i] << " "; cout << endl;        solve (1, n);        //for (int i = 1; i <= n; i++) cout << f[i] << " "; cout << endl;        long long ans = f[n];         long long tmp = qpow (n, n-2) * qpow (m, n-1) % mod;        ans = (ans - tmp + mod) % mod;        printf ("Case #%d: %lld\n", ++kase, ans);    }    return 0;}

NTT：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <map>#include <vector>#include <stack>using namespace std;#define mod 152076289#define G 106#define maxn 40005long long x1[maxn], x2[maxn];long long qpow (long long a, long long b) {      long long ret=1;      while (b) {          if (b&1) ret = (ret*a)%mod;          a = (a*a)%mod;          b >>= 1;      }      return ret;  }  void change (long long *y, int len) {    int i, j, k;    for(i = 1, j = len / 2; i < len - 1; i++) {        if (i < j) swap(y[i], y[j]);        k = len / 2;        while (j >= k) {            j -= k;            k /= 2;        }        if (j < k) j += k;    }}void ntt (long long *y, int len, int on) {    change (y, len);    int id = 0;    for(int h = 2; h <= len; h <<= 1) {        id++;        long long wn = qpow (G, (mod - 1) / (1<<id));        for(int j = 0; j < len; j += h) {            long long w = 1;            for(int k = j; k < j + h / 2; k++) {                long long u = y[k] % mod;                long long t = w * (y[k + h / 2] % mod) % mod;                y[k] = (u + t) % mod;                y[k + h / 2] = ((u - t) % mod + mod) % mod;                w = w * wn % mod;            }        }    }    if (on == -1) {        for (int i = 1; i < len / 2; i++)            swap (y[i], y[len - i]);        long long inv = qpow(len, mod - 2);        for(int i = 0; i < len; i++)            y[i] = y[i] % mod * inv % mod;    }}long long n, m;long long f[maxn], g[maxn], h[maxn];//n个点的连通图 n个点的图 n个点的树long long c[maxn], fac[maxn], rev[maxn];void solve (int l, int r) {    if (l == r) {        f[l] += g[l];        f[l] %= mod;        return;    }    int mid = (l+r) >> 1;    solve (l, mid);    int len = 1;    while (len <= r-l+1) {len <<= 1;}    for (int i = 0; i < len; i++) {        x1[i] = x2[i] = 0;    }    for (int i = l; i <= mid; i++) {        x1[i-l] = f[i]*rev[i-1]%mod;    }    for (int i = 1; i <= r-l; i++) {        x2[i-1] = g[i]*rev[i]%mod;    }    ntt (x1, len, 1), ntt (x2, len, 1);    for (int i = 0; i < len; i++) x1[i] = x1[i] * x2[i] % mod;    ntt (x1, len, -1);    for (int i = mid+1; i <= r; i++) {        f[i] -= x1[i-l-1] %mod *fac[i-1] %mod;        (f[i] += mod) %= mod;    }    solve (mid+1, r);}int main () {    int t, kase = 0;    fac[0] = 1;    for (int i = 1; i < maxn; i++) fac[i] = fac[i-1] * i % mod;    rev[maxn-1] = qpow (fac[maxn-1], mod - 2);    for (int i = maxn-2; i >= 0; i--) rev[i] = rev[i+1] * (i+1) % mod;    scanf ("%d", &t);    while (t--) {        scanf ("%lld%lld", &n, &m);        memset (f, 0, sizeof f);        for (int i = 1; i <= n; i++) {            g[i] = qpow (1+m, i*(i-1)/2);        }        solve (1, n);        long long ans = f[n];         long long tmp = qpow (n, n-2) * qpow (m, n-1) % mod;        ans = (ans - tmp + mod) % mod;        printf ("Case #%d: %lld\n", ++kase, ans);    }    return 0;}