[期望 状压DP 子集DP 多项式] BZOJ 3925 [Zjoi2015]地震后的幻想乡

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
typedef __float128 ld;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

struct Poly{
int n; vector<ll> a;
Poly(){ }
Poly(int in){
if (in>=0) { n=in; a.resize(n+1); for (int i=0;i<=n;i++) a[i]=0; }
if (in==-1) n=0,a.resize(1),a[0]=1;
}
ll &operator [](int x){
return a[x];
}
friend Poly operator + (const Poly &A,const Poly &B){
Poly ret=Poly(max(A.n,B.n));
for (int i=0;i<=A.n;i++) ret[i]+=A.a[i];
for (int i=0;i<=B.n;i++) ret[i]+=B.a[i];
return ret;
}
friend Poly operator - (const Poly &A,const Poly &B){
Poly ret=Poly(max(A.n,B.n));
for (int i=0;i<=A.n;i++) ret[i]+=A.a[i];
for (int i=0;i<=B.n;i++) ret[i]-=B.a[i];
return ret;
}
friend Poly operator * (const Poly &A,const Poly &B){
Poly ret=Poly(A.n+B.n);
for (int i=0;i<=A.n;i++) for (int j=0;j<=B.n;j++) ret[i+j]+=A.a[i]*B.a[j];
return ret;
}
};

const int N=15;

int n,m,a
;
Poly pow[N*N],f[1<<10],Ans;
int cnt[1<<10],bin[1<<10];

#define low(x) (bin[(x)&-(x)])

int main(){
int iu,iv;
freopen("mst.in","r",stdin);
freopen("mst.out","w",stdout);
read(n); read(m);
for (int i=1;i<(1<<n);i++) cnt[i]=cnt[i^(i&-i)]+1;
for (int i=1;i<=n;i++) bin[1<<(i-1)]=i;
for (int i=1;i<=m;i++)
read(iu),read(iv),a[iu]+=1<<(iv-1),a[iv]+=1<<(iu-1);
pow[0]=Poly(-1); pow[1]=Poly(1),pow[1][0]=1,pow[1][1]=-1;
for (int i=2;i<=m;i++) pow[i]=pow[i-1]*pow[1];
for (int i=1;i<(1<<n);i++){
f[i]=Poly(-1);
int rest=i^(1<<(low(i)-1));
for (int j=rest;j;(--j)&=rest){
int cot=0;
for (int t=i^j;t;t-=t&-t)
cot+=cnt[a[low(t)]&j];
f[i]=f[i]-f[i^j]*pow[cot];
}
}
Ans=Poly(-1)-f[(1<<n)-1];
ld ans=0;
for (int i=0;i<=Ans.n;i++) ans+=(ld)Ans[i]/(i+1);
printf("%.6lf\n",(double)ans);
return 0;
}