有标号的二分图计数 [生成函数 多项式]

有标号的二分图计数

题目也在COGS上

[HZOI 2015]有标号的二分图计数 I

[HZOI 2015]有标号的二分图计数 II

[HZOI 2015]有标号的二分图计数 III

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I

求n个点的二分图(可以不连通)的个数。(n le 10^5)

其中二分图进行了黑白染色，两个二分图不同：边不同 或 点的颜色不同

水题啊，只有黑白之间连边。

[sum_{k=0}^n inom{n}{k} 2^{k(n-k)} ]

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II

求n个点的二分图(可以不连通)的个数。(n le 10^5)

不能简单的除以2，问题在于有的黑白之间不连边

i个连通块，贡献就是(2^i)

DP (f(n,i))表示n个点i个连通块的二分图个数，(O(n^3))

考虑生成函数！

(S(x))表示上道题，(F(x))表示本题

还是不好做，因为都与连通块有关，引入(H(x))表示单个连通块！

[S(x) = sum_{i ge 0} frac{2^i cdot H(x)^i}{i!} \ F(x) = sum_{i ge 0} frac{H(x)^i}{i!} = sqrt{S(x)}\ ]

多项式开根即可

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III

求n个点的二分图(必须连通)的个数。(n le 10^5)

就是(H(x))

就是(frac{1}{2} ln S(x))

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Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 1e5+5, P = 998244353;
inline int read() {
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

ll Pow(ll a, int b) {
	ll ans = 1;
	for(; b; b>>=1, a=a*a%P)
		if(b&1) ans=ans*a%P;
	return ans;
}
int n;
ll inv[N], fac[N], facInv[N];
inline ll C(int n, int m) {return fac[n] * facInv[m] %P * facInv[n-m] %P;}
int main() {
	freopen("QAQ_bipartite_one.in", "r", stdin);
	freopen("QAQ_bipartite_one.out", "w", stdout);
	//freopen("in", "r", stdin);
	n = read();
	inv[1] = fac[0] = facInv[0] = 1;
	for(int i=1; i<=n; i++) {
		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
		fac[i] = fac[i-1] * i %P;
		facInv[i] = facInv[i-1] * inv[i] %P;
	}
	ll ans = 0;
	for(int k=0; k<=n; k++) ans = (ans + C(n, k) * Pow(2, (ll) k * (n-k) % (P-1))) %P;
	printf("%lld
", ans);
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = (1<<18) + 5, P = 998244353, qr2 = 116195171, inv2 = (P+1)/2;
inline int read() {
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

ll Pow(ll a, int b) {
	ll ans = 1;
	for(; b; b >>= 1, a = a * a %P)
		if(b & 1) ans = ans * a %P;
	return ans;
}

namespace fft {
	int rev[N];
	void dft(int *a, int n, int flag) { 
		int k = 0; while((1<<k) < n) k++;
		for(int i=0; i<n; i++) {
			rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
			if(i < rev[i]) swap(a[i], a[rev[i]]);
		}
		for(int l=2; l<=n; l<<=1) {
			int m = l>>1;
			ll wn = Pow(3, flag == 1 ? (P-1)/l : P-1-(P-1)/l);
			for(int *p = a; p != a+n; p += l) 
				for(int k=0, w=1; k<m; k++, w = w*wn%P) {
					int t = (ll) w * p[k+m] %P;
					p[k+m] = (p[k] - t + P) %P;
					p[k] = (p[k] + t) %P;
				}
		}
		if(flag == -1) {
			ll inv = Pow(n, P-2);
			for(int i=0; i<n; i++) a[i] = a[i] * inv %P;
		}
	}

	void sqr(int *a, int n) {
		dft(a, n, 1);
		for(int i=0; i<n; i++) a[i] = (ll) a[i] * a[i] %P;
		dft(a, n, -1);
	}

	void inverse(int *a, int *b, int l) {
		static int t[N];
		if(l == 1) {b[0] = Pow(a[0], P-2); return;}
		inverse(a, b, l>>1);
		int n = l<<1;
		for(int i=0; i<l; i++) t[i] = a[i], t[i+l] = 0; 
		dft(t, n, 1); dft(b, n, 1);
		for(int i=0; i<n; i++) b[i] = (ll) b[i] * (2 - (ll) t[i] * b[i] %P + P) %P;
		dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;
	}

	void sqrt(int *a, int *b, int l) {
		static int t[N], ib[N];
		if(l == 1) {b[0] = 1; return;}
		sqrt(a, b, l>>1);
		int n = l<<1;
		for(int i=0; i<l; i++) t[i] = a[i], t[i+l] = ib[i] = ib[i+l] = 0;
		inverse(b, ib, l);
		dft(t, n, 1); dft(b, n, 1); dft(ib, n, 1);
		for(int i=0; i<n; i++) b[i] = (ll) inv2 * (b[i] + (ll) t[i] * ib[i] %P) %P;
		dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;
	}
}

int n, a[N], f[N], len;
ll inv[N], fac[N], facInv[N], mi[N];
int main() {
	//freopen("in", "r", stdin);
	freopen("QAQ_bipartite_two.in", "r", stdin);
	freopen("QAQ_bipartite_two.out", "w", stdout);
	n = read();
	len = 1; while(len <= n) len <<= 1;
	inv[1] = fac[0] = facInv[0] = 1;
	for(int i=1; i<=n; i++) {
		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
		fac[i] = fac[i-1] * i %P;
		facInv[i] = facInv[i-1] * inv[i] %P;
	}

	for(int i=0; i<=n; i++) a[i] = facInv[i] * Pow(Pow(qr2, (ll) i * i %(P-1) ), P-2) %P;
	fft::sqr(a, len<<1);
	for(int i=0; i<=n; i++) a[i] = a[i] * Pow(qr2, (ll) i * i %(P-1)) %P;
	fft::sqrt(a, f, len);
	printf("%lld
", (ll) f[n] * fac[n] %P);
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = (1<<18) + 5, P = 998244353, qr2 = 116195171, inv2 = (P+1)/2;
inline int read() {
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

ll Pow(ll a, int b) {
	ll ans = 1;
	for(; b; b >>= 1, a = a * a %P)
		if(b & 1) ans = ans * a %P;
	return ans;
}

ll inv[N], fac[N], facInv[N];
namespace fft {
	int rev[N];
	void dft(int *a, int n, int flag) { 
		int k = 0; while((1<<k) < n) k++;
		for(int i=0; i<n; i++) {
			rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));
			if(i < rev[i]) swap(a[i], a[rev[i]]);
		}
		for(int l=2; l<=n; l<<=1) {
			int m = l>>1;
			ll wn = Pow(3, flag == 1 ? (P-1)/l : P-1-(P-1)/l);
			for(int *p = a; p != a+n; p += l) 
				for(int k=0, w=1; k<m; k++, w = w*wn%P) {
					int t = (ll) w * p[k+m] %P;
					p[k+m] = (p[k] - t + P) %P;
					p[k] = (p[k] + t) %P;
				}
		}
		if(flag == -1) {
			ll inv = Pow(n, P-2);
			for(int i=0; i<n; i++) a[i] = a[i] * inv %P;
		}
	}

	void sqr(int *a, int n) {
		dft(a, n, 1);
		for(int i=0; i<n; i++) a[i] = (ll) a[i] * a[i] %P;
		dft(a, n, -1);
	}

	void inverse(int *a, int *b, int l) {
		static int t[N];
		if(l == 1) {b[0] = Pow(a[0], P-2); return;}
		inverse(a, b, l>>1);
		int n = l<<1;
		for(int i=0; i<l; i++) t[i] = a[i], t[i+l] = 0; 
		dft(t, n, 1); dft(b, n, 1);
		for(int i=0; i<n; i++) b[i] = (ll) b[i] * (2 - (ll) t[i] * b[i] %P + P) %P;
		dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;
	}

    void ln(int *a, int *b, int l) {
        static int da[N], ia[N];
        int n = l<<1;
        for(int i=0; i<n; i++) da[i] = ia[i] = 0;
        for(int i=0; i<l-1; i++) da[i] = (ll) (i+1) * a[i+1] %P;
        inverse(a, ia, l);
        dft(da, n, 1); dft(ia, n, 1);
        for(int i=0; i<n; i++) b[i] = (ll) da[i] * ia[i] %P;
        dft(b, n, -1);
        for(int i=l-1; i>0; i--) b[i] = (ll) inv[i] * b[i-1] %P; b[0] = 0;
        for(int i=l; i<n; i++) b[i] = 0;
    }
}

int n, a[N], f[N], len;
int main() {
	//freopen("in", "r", stdin);
	freopen("QAQ_bipartite_thr.in", "r", stdin);
	freopen("QAQ_bipartite_thr.out", "w", stdout);
	n = read();
	len = 1; while(len <= n) len <<= 1;
	inv[1] = fac[0] = facInv[0] = 1;
	for(int i=1; i<=len; i++) {
		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;
		fac[i] = fac[i-1] * i %P;
		facInv[i] = facInv[i-1] * inv[i] %P;
	}

	for(int i=0; i<=n; i++) a[i] = facInv[i] * Pow(Pow(qr2, (ll) i * i %(P-1) ), P-2) %P;
	fft::sqr(a, len<<1);
	for(int i=0; i<=n; i++) a[i] = a[i] * Pow(qr2, (ll) i * i %(P-1)) %P;
	fft::ln(a, f, len);
	printf("%lld
", f[n] * fac[n] %P * inv2 %P);
}