BZOJ 4455

树的点到图的点是双射

枚举哪些点可以映射到

然后dp容斥

复杂度 $2^n*n^3$

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=(a),i##_end=(b);i<=i##_end;++i)
#define For(i,a,b) for(int i=(a),i##_end=(b);i<i##_end;++i)
#define per(i,a,b) for(int i=(b),i##_st=(a);i>=i##_st;--i)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define dbg(x) cerr<<#x" = "<<x<<endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Es(x,i) for(Edge *i=G[x];i;i=i->nxt)
typedef long long ll;
typedef pair<int,int> pii;
const int inf=~0u>>1,mod=1e9+7;
inline int rd() {
    int x,c,f=1;while(!isdigit(c=getchar()))f=c!='-';x=c-'0';
    while(isdigit(c=getchar()))x=x*10+c-'0';return f?x:-x;
}
const int N=18;
struct Edge{int v;Edge*nxt;}pl[N*N],*cur=pl,*G[N];
inline void ins(int u,int v){*cur=(Edge){v,G[u]},G[u]=cur++;}
char g[N][N];
int n,m,a[N],tot;
ll f[N][N];
inline void Dp(int x,int fa=-1){
	Es(x,i)if(i->v!=fa){
		Dp(i->v,x);
	}
	For(i,0,tot){
		f[x][i]=1;
		Es(x,p)if(p->v!=fa){
			ll t=0;
			For(j,0,tot)if(g[a[i]][a[j]]){
				t+=f[p->v][j];
			}
			f[x][i]*=t;
		}
	}
}
int main(){
#ifdef flukehn
	freopen("ex_star2.in","r",stdin);
#endif
	n=rd(),m=rd();
	rep(i,1,m){
		int u=rd()-1,v=rd()-1;
		g[u][v]=g[v][u]=1;
	}
	ll ans=0;
	For(i,1,n){
		int u=rd()-1,v=rd()-1;
		ins(u,v),ins(v,u);
	}
	For(i,1,1<<n){
		tot=0;
		For(j,0,n)if(i>>j&1)a[tot++]=j;
		Dp(0);
		ll p=0;
		For(j,0,tot)p+=f[0][j];
		ans+=((tot&1)?-1:1)*p;
	}
	if(ans<0)ans=-ans;
	cout<<ans<<endl;
}