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
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。
\[\sum_{k=0}^n \binom{n}{k} 2^{k(n-k)}
\]
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
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\(\frac{1}{2} \ln S(x)\)
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\n", 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\n", (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\n", f[n] * fac[n] %P * inv2 %P);
}
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