PKUWC2018 随机算法
考虑最暴力的状压DP，每个点有3种状态：在独立集中，考虑了但不在独立集中，没考虑。
然后就可以$O(n3^n)$DP了。
但是显然是过不了的哇，所以考虑优化。
发现如果一个点已经被选在独立集中了，那么与它相邻的点一定不在独立集中，并且这些点不管什么时候加入都不在独立集中。
所以当我们考虑把一个点放进独立集中，所有与它相邻并且没考虑过的点都可以安排在后面的任意一个位置对答案都不影响。
那么我们认为考虑了一个点，当且仅当这个点在独立集中或与独立集中的点联通，那么枚举下一个考虑的点，这个点都可以加入独立集。
然后dp[t][S]表示独立集大小为t，考虑点集为S的方案数 就可以了。时间复杂度$O(n^22^n)$
/*
Author: CNYALI_LK
LANG: C++
PROG: 2540.cpp
Mail: cnyalilk@vip.qq.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 signed inf=0x3f3f3f3f;
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 void read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}
inline void read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}
inline void read (char &x) {
x=gc();
}
inline void read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r');
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
}
template<typename A,typename ...B>
inline void read(A &x,B &...y){
read(x);read(y...);
}
// print a signed integer
inline void 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 --]);
}
inline void 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 --]);
}
inline void write (char x) {
putc(x);
}
inline void write(const char *x){
while(*x){putc(*x);++x;}
}
inline void write(char *x){
while(*x){putc(*x);++x;}
}
template<typename A,typename ...B>
inline void write(A x,B ...y){
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;
const int p=998244353;
int sum(int a,int b){return (a+=b)>=p?a-p:a;}
int n,m,u,v,mx,a[25],f[25][1<<20],A[25][25],deg[25],fulls;
bool vis[25][1<<20];
void DFS(int x,int y){
if(vis[x][y])return;
vis[x][y]=1;
if(y+1 == 1<<n){f[x][y]=x==mx;return;}
for(int i=0;i<n;++i)if(!(y&1<<i)){
int xy=y|(1<<i)|a[i];
DFS(x+1,xy);
f[x][y]=sum(f[x][y],(ll)f[x+1][xy]*A[n-__builtin_popcount(y)-1][__builtin_popcount(a[i]&(fulls^y))]%p);
}
}
int fpm(int a,int b){int c=1;for(;b;b>>=1,a=(ll)a*a%p)if(b&1)c=(ll)c*a%p;return c;}
int main(){
#ifdef cnyali_lk
freopen("2540.in","r",stdin);
freopen("2540.out","w",stdout);
#endif
read(n,m);
for(int i=0;i<=n;++i){
A[i][0]=1;
for(int j=1;j<=i;++j)A[i][j]=(ll)A[i][j-1]*(i-j+1)%p;
}
fulls=(1<<n)-1;
for(;m;--m){
read(u,v);
--u;--v;
a[u]|=1<<v;
a[v]|=1<<u;
++deg[u];
++deg[v];
}
mx=0;
int s,ok;
for(int i=1;i<1<<n;++i){
ok=1,s=__builtin_popcount(i);
if(s<mx)continue;
for(int j=0;j<n;++j)if((i&(1<<j)) && (i&a[j])){ok=0;break;}
if(ok)mx=s;
}
DFS(0,0);
s=1;
for(int i=1;i<=n;++i)s=(ll)s*i%p;
write((ll)f[0][0]*fpm(s,p-2)%p,'\n');
return 0;
}