hihocoder #1071 : 小玩具

闻所未闻的$dp$神题（我不会的题）

令$f[S][i]$表示子集状态为$S$，且$S$中最大联通块恰好为$i$的方案数

考虑转移，我们枚举$S$中最小的元素$v$来转移，这样就能不重

$f[S][i] = \sum\limits_{T \in S \;and\;v \in T} f[T][...] * C[S \wedge T]$

由于这么递归转移不好确定后面的状态，因此我们可以递推转移，在代码中有所体现

$C[S]$表示将$S$联通的方案数

我们考虑容斥，用全集减去所有不联通的方案数，我们考虑枚举最小点$v$所在的集合

之后转移时$C[S] = 2^{E[S]} - \sum\limits_{T \in s \;ans\;v\; \in T} C[T] * 2^{E[S \wedge T]}$

其中，$E[S]$表示处于$S$内部的边的方案数

复杂度$O(3^n * n)$

#include <set>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define re register
    #define de double
    #define le long double
    #define ri register int
    #define ll long long
    #define sh short
    #define pii pair<int, int>
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)
    #define gc getchar
    inline int read() {
        int p = , w = ; char c = gc();
        while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
        while(c >= '' && c <= '') p = p *  + c - '', c = gc();

        return p * w;
    }
    int wr[], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('');
        if(o < ) o = -o, pc('-');
        while(o) wr[++ rw] = o % , o /= ;
        while(rw) pc(wr[rw --] + '');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; }
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b,  : ; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b,  : ; }
}
using namespace std;
using namespace remoon;

#define sid 17
#define mod 1000000007

inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; }
inline void dec(int &a, int b) { a -= b; if(a < ) a += mod; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }

int pc[];
int u[], v[];
int E[( << ) + ], C[( << ) + ];
int f[( << ) + ][sid];
int n, m;

int main() {

    n = read(); m = read(); pc[] = ;
    rep(i, , ) pc[i] = mul(pc[i - ], );
    rep(i, , m) u[i] = read(), v[i] = read();

    rep(S, , ( << n) - ) rep(i, , m)
    if((S & ( << u[i] - )) && (S & ( << v[i] - ))) ++ E[S];

    C[] = ;
    rep(S, , ( << n) - ) {
        int res = pc[E[S]], mi = -;
        rep(i, , n) if(S & ( << i - )) { mi = i; break; }
        for(ri T = S & (S - ); T; T = (T - ) & S)
        if(T & ( << mi - )) dec(res, mul(C[T], pc[E[S ^ T]]));
        C[S] = res;
    }

    f[][] = ;
    rep(S, , ( << n) - ) rep(j, , n) if(f[S][j]) {
        int mi = -;
        rep(i, , n) if(!(S & ( << i - ))) { mi = i; break; }
        if(mi == -) continue;
        int D = (( << n) - ) ^ (S | ( << mi - ));
        for(ri T = D; ; T = (T - ) & D) {
            int st = __builtin_popcount(T | ( << mi - ));
            inc(f[S | ( << mi - ) | T][max(j, st)], mul(f[S][j], C[( << mi - ) | T]));
            if(!T) break;
        }
    }

    rep(i, , n)
    write(f[( << n) - ][i]);

    return ;
}