【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;

}

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【51nod】1634 刚体图

【51nod】1634 刚体图

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