CometOJ Round #11 F

题意

给定一张n点m边的图,每条边$(u,v)$有$\frac{1}{3}$的概率$u$指向$v$,有$\frac{1}{3}$的概率从$v$指向$u$,还有$\frac{1}{3}$的概率消失,求这张图是DAG的概率。

$n\le 20,4s$

题解

显然可以转化求为每条边可以u指v,可以v指u,可以消失,最后成DAG的方案数。

设$f_S$为$S$中形成DAG的概率。

枚举S中入度为0的点集T($T\not=\varnothing$),显然一个入度为零的点集会被每个非空子集算一次。

凑系数$a_{|T|}$满足$\sum_{i=1}^{|T|} \binom{|T|}{i}a_{i}=1$,发现$a_{|T|}=(-1)^{|T|-1}$。

那么$f_S=\sum_{T\subset S,T\not=\varnothing}f_{S-T}2^{e_{T,S-T}}$,其中$e_{T,S-T}$表示$T$到$S-T$的边数,也就是$e_S-e_T-e_{S-T}$其中$e_S$为$S$内部的边数。那么$2^{e_{T,S-T}}=\frac{2^{e_S}}{2^{e_T}2^{e_{S-T}}}$

$$\frac{f_S}{2^{e_S}}=\sum_{T\subset S,T\not=\varnothing}\frac{f_{S-T}}{2^{e_{S-T}}}\frac{(-1)^{|T|-1}}{2^{e_T}}$$

可以子集卷积。

注意:子集卷积中,集合并卷积时有可能$f_{s,|S|}$中$s\lt |S|$,但是这样对答案没有影响,不需要iFMT一下消除这些项。

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/*
Author: QAQ Automaton
Lang: C++
Prog: F.cpp
Mail: lk@qaq-am.com
Blog: https://www.qaq-am.com/
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const int SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// print the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed integer
inline bool read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}

inline bool read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
return 1;
}
inline bool read (char &x) {
x=gc();
return x!=EOF;
}
inline bool read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
while(!(*x=='\n'||*x==' '||*x=='\r'||*x==EOF))*(++x)=gc();
*x=0;
return 1;
}
template<typename A,typename ...B>
inline bool read(A &x,B &...y){
return read(x)&&read(y...);
}
// print a signed integer
inline bool write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}

inline bool write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
return 0;
}
inline bool write (char x) {
putc(x);
return 0;
}
inline bool write(const char *x){
while(*x){putc(*x);++x;}
return 0;
}
inline bool write(char *x){
while(*x){putc(*x);++x;}
return 0;
}
template<typename A,typename ...B>
inline bool write(A x,B ...y){
return write(x)||write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
int inf;
struct _init_{
_init_(){
memset(&inf,0x3f,sizeof(inf));
}
};
_init_ ___INIT__;
int e[1<<20];
int t[20];
int is[1<<20];
const int p=998244353,i2=(p+1)>>1,i3=(p+1)/3;
int p2[405],p3[405];
int f[21][1<<20],g[21][1<<20],pc[1<<20];
void FMT(int *f,int n){
for(int i=1;i<n;i<<=1)
for(int j=0;j<n;++j)if(!(j&i))f[j|i]=(f[j|i]+f[j])%p;
}
void iFMT(int *f,int n){
for(int i=1;i<n;i<<=1)
for(int j=0;j<n;++j)if(!(j&i))f[j|i]=(f[j|i]+p-f[j])%p;
}
signed main(){
#ifdef QAQAutoMaton
freopen("F.in","r",stdin);
freopen("F.out","w",stdout);
#endif
int n,m,u,v;
read(n,m);
p2[0]=p3[0]=1;
for(int i=1;i<=m;++i){
p2[i]=(ll)p2[i-1]*i2%p;
p3[i]=(ll)p3[i-1]*2*i3%p;
}
for(int i=1;i<=m;++i){
read(u,v);
--u;--v;
t[u]|=1<<v;
t[v]|=1<<u;
}
for(int i=0;i<n;++i)is[1<<i]=i;
for(int i=1;i<1<<n;++i)pc[i]=pc[i&(i-1)]+1;
for(int i=1;i<1<<n;++i){
e[i]=e[i&(i-1)]+pc[t[is[i&(-i)]]&i];
}
for(int i=1;i<1<<n;++i)g[pc[i]][i]=(ll)(pc[i]&1?1:p-1)*p2[e[i]]%p;
for(int i=0;i<=n;++i)FMT(g[i],1<<n);
f[0][0]=1;
FMT(f[0],1<<n);
for(int i=1;i<=n;++i){
for(int j=0;j<i;++j)for(int k=0;k<1<<n;++k)f[i][k]=(f[i][k]+(ll)f[j][k]*g[i-j][k])%p;
}
iFMT(f[n],1<<n);

write((ll)f[n][(1<<n)-1]*p3[m]%p,'\n');
/* f[0]=1;
for(int i=1;i<1<<n;++i)for(int j=i;j;j=(j-1)&i){
f[i]=(f[i]+f[i-j]*(__builtin_popcount(j)&1?1:p-1)%p*p2[e[j]])%p;
}
write(f[(1<<n)-1]*p3[m]%p,'\n');*/
return 0;
}
# 数学

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