【LOJ #2540】「PKUWC2018」随机算法（状压DP）

传送门

设$f[i][s]$为考虑排列里第$i$个点，当前独立集为$s$的方案数
暴力枚举下一个点转移即可
可以加入独立集的一定还没考虑过

复杂度$O(n^22^n)$但是转移次数很少
跑的飞快

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define poly vector<int>
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=998244353;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return r>=mod?(r%mod):r;}
inline void Add(int&a,int b){a+=b,a>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int N=21,M=(1<<20)|5;
int f[N][M];
int conc[N];
int n,m,sta,cnt[M];
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	n=read(),m=read(),sta=1<<n;
	for(int i=1;i<=m;i++){
		int u=read()-1,v=read()-1;
		conc[u]|=(1<<v),conc[v]|=(1<<u);
	}
	for(int i=1;i<=sta;i++)cnt[i]=cnt[i>>1]+(i&1);
	f[0][0]=1;
	for(int i=1;i<=n;i++){
		for(int s=0;s<sta;s++)if(f[i-1][s]){
			int cnt=n-i+1;
			for(int j=0;j<n;j++)if(!(s&(1<<j))){
				if(!(s&conc[j]))Add(f[i][s|(1<<j)],f[i-1][s]),cnt--;
			}
			Add(f[i][s],mul(f[i-1][s],cnt));
		}
	}
	int mx=0,res=0;
	for(int s=0;s<sta;s++){
		if(cnt[s]>mx&&f[n][s])mx=cnt[s],res=0;
		if(cnt[s]==mx)Add(res,f[n][s]);
	}
	int mt=1;
	for(int i=2;i<=n;i++)Mul(mt,i);
	cout<<mul(res,Inv(mt));
}