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

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

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$C[S] = 2^{E[S]} - sumlimits_{T in s ;ans;v; in T} C[T] * 2^{E[S wedge T]}$

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 = 0, w = 1; char c = gc();
        while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
        while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
        
        return p * w;
    }
    int wr[50], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '
') {
        if(!o) pc('0');
        if(o < 0) o = -o, pc('-');
        while(o) wr[++ rw] = o % 10, o /= 10;
        while(rw) pc(wr[rw --] + '0');
        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, 1 : 0; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; }
}
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 < 0) a += mod; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }

int pc[2555];
int u[555], v[555];
int E[(1 << 16) + 5000], C[(1 << 16) + 5000];
int f[(1 << 16) + 5000][sid];
int n, m;

int main() {
    
    n = read(); m = read(); pc[0] = 1;
    rep(i, 1, 1000) pc[i] = mul(pc[i - 1], 2);
    rep(i, 1, m) u[i] = read(), v[i] = read();
    
    rep(S, 0, (1 << n) - 1) rep(i, 1, m) 
    if((S & (1 << u[i] - 1)) && (S & (1 << v[i] - 1))) ++ E[S];
        
    C[0] = 1;
    rep(S, 1, (1 << n) - 1) {
        int res = pc[E[S]], mi = -1; 
        rep(i, 1, n) if(S & (1 << i - 1)) { mi = i; break; }
        for(ri T = S & (S - 1); T; T = (T - 1) & S)
        if(T & (1 << mi - 1)) dec(res, mul(C[T], pc[E[S ^ T]]));
        C[S] = res;
    }
        
    f[0][0] = 1;
    rep(S, 0, (1 << n) - 1) rep(j, 0, n) if(f[S][j]) {
        int mi = -1;
        rep(i, 1, n) if(!(S & (1 << i - 1))) { mi = i; break; }
        if(mi == -1) continue;
        int D = ((1 << n) - 1) ^ (S | (1 << mi - 1));
        for(ri T = D; ; T = (T - 1) & D) {
            int st = __builtin_popcount(T | (1 << mi - 1));
            inc(f[S | (1 << mi - 1) | T][max(j, st)], mul(f[S][j], C[(1 << mi - 1) | T]));
            if(!T) break;
        }
    }
    
    rep(i, 1, n) 
    write(f[(1 << n) - 1][i]);
    
    return 0;
}