设f(n)为所求答案

g(n)为n个顶点的非联通图

则f(n) + g(n) = h(n) = 2^(n * (n - 1) / 2)

其中h(n)是n个顶点的联图的个数

这样计算

先考虑1所在的连通分量包含哪些顶点

假设该连通分量有k个顶点

就有C(n - 1, k - 1)种集合

确定点集后，所在的连通分量有f(k)种情况。其他连通分量有 h(n - k)种情况

因此有递推公式。g(n) = sum{ C(n - 1, k - 1) * f(k) * h(n - k)} 其中k = 1,2...n-1

注意每次计算出g(n)后立刻算出f(n)

 

import java.math.BigInteger;import java.util.Scanner;public class Main {		public static void main(String[] args) {		BigInteger two[] = new BigInteger [55];		two[0] = BigInteger.ONE;		for(int i = 1; i <= 50; i++)			two[i] = two[i - 1].multiply(BigInteger.valueOf(2));		BigInteger h[] = new BigInteger [55];		for(int i = 1; i <= 50; i++){			int a = i;			int b = i - 1;			if(a % 2 == 0) a/= 2;			else b /= 2;			h[i] = BigInteger.ONE;			for(int j = 0; j < a; j++) {				h[i] = h[i].multiply(two[b]);			}		}		BigInteger C[][] = new BigInteger[55][55];		C[0][0] = BigInteger.ONE;		for(int i = 0; i <= 50; i++){			C[i][0] = C[i][i] = BigInteger.ONE;			for(int j = 1; j < i; j++){				C[i][j] = C[i - 1][j].add(C[i - 1][j - 1]);			}		}		BigInteger f[] = new BigInteger[55];		BigInteger g[] = new BigInteger[55];		f[1] = BigInteger.ONE;		for(int i = 2; i <= 50; i++) {			g[i] = BigInteger.ZERO;			for(int j = 1; j < i; j++) {				g[i] = g[i].add(C[i - 1][j - 1].multiply(f[j]).multiply(h[i - j]));			}			f[i] = h[i].subtract(g[i]);		}		int n;		Scanner cin = new Scanner(System.in);		while(cin.hasNext()){			n = cin.nextInt();			if(n == 0) break;			System.out.println(f[n]);		}	}}