bzoj 2560 串珠子

大意: n个点, $i$和$j$之间可以连$c_{ij}$种无向边, 求连成连通图的方案数.

设$f_i$为状态$i$时连通图方案, $g_i$为状态$i$时所有方案.

有$f[i]=g[i]-sumlimits_{j}f[j]g[iwedge j]$, j为包含i的最低位的所有i的子集

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 16;
int n;
int a[N][N], f[1<<N], g[1<<N];

int main() {
	scanf("%d", &n);
	REP(i,0,n-1) REP(j,0,n-1) scanf("%d",a[i]+j);
	int mx = (1<<n)-1;
	REP(k,0,mx) {
		f[k] = 1;
		REP(i,0,n-2) if (k>>i&1) {
			REP(j,i+1,n-1) if (k>>j&1) {
				f[k] = (ll)f[k]*(a[i][j]+1)%P;
			}
		}
		g[k] = f[k];
		for (int i=k^k&-k,j=i; j; (--j)&=i) {
			f[k] = (f[k]-(ll)g[j]*f[k^j])%P;
		}
	}
	int ans = f[mx];
	if (ans<0) ans+=P;
	printf("%d
", ans);
}