【51nod】1634 刚体图

给一个左边n个点右边m个点二分图求合法的连通图个数，每条边选了之后会带来价值乘2的贡献

类似城市规划那道题的计数

设\(g[i][j]\)为左边\(i\)个点，右边\(j\)个点的图有多少个（就是边随便连）

\(f[i][j]\)为左边\(i\)个点右边\(j\)个点的连通图有多少个

然后枚举和左边第一个点连通的联通块是几个左边点，几个右边点

答案可以认为是

\(f[i][j] = g[i][j] - \sum_{k = 0}^{i - 1}\sum_{h = 0}^{j} \binom{i - 1}{k}\binom{j}{h} f[i - k][j - h] \times g[k][h]\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,M;
int f[15][15],g[15][15],C[105][105];
int fac[15],invfac[15];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
    if(c & 1) res = mul(res,t);
    t = mul(t,t);
    c >>= 1;
    }
    return res;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    for(int i = 0 ; i <= 100 ; ++i) {
    C[i][0] = 1;
    for(int j = 1 ; j <= i ; ++j) {
        C[i][j] = inc(C[i - 1][j],C[i - 1][j - 1]);
    }
    }
    for(int i = 0 ; i <= 10 ; ++i) {
    for(int j = 0 ; j <= 10 ; ++j) {
        for(int k = 0 ; k <= i * j ; ++k) {
        update(g[i][j],mul(C[i * j][k],fpow(2,k)));
        }
    }
    }
    
    
    while(scanf("%d%d",&N,&M) != EOF) {
    memset(f,0,sizeof(f));
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = 0 ; j <= M ; ++j) {
        f[i][j] = g[i][j];
        for(int k = 0 ; k < i ; ++k) {
            for(int h = 0 ;h <= j ; ++h) {
            if(!(h + k)) continue;
            int c = mul(C[j][h],C[i - 1][k]);
            c = mul(c,mul(f[i - k][j - h],g[k][h]));
            update(f[i][j],MOD - c);
            }
        }
        }
    }
    out(f[N][M]);enter;
    }
    return 0;
}