poj1737 组合dp

http://poj.org/problem?id=1737

Description

An undirected graph is a set V of vertices and a set of E∈{V*V} edges.An undirected graph is connected if and only if for every pair (u,v) of vertices,u is reachable from v. 
You are to write a program that tries to calculate the number of different connected undirected graph with n vertices. 
For example,there are 4 different connected undirected graphs with 3 vertices. 

Input

The input contains several test cases. Each test case contains an integer n, denoting the number of vertices. You may assume that 1<=n<=50. The last test case is followed by one zero.

Output

For each test case output the answer on a single line.

Sample Input

1
2
3
4
0

Sample Output

1
1
4
38

/**
poj1737 组合dp
题目大意：给定节点数n，问由这些节点组成的连通图有多少种
解题思路：从n个点所有的连接方法中删去那些不能构成连通图的连法.然后想法就是用f[i]记录i个点的连通图数目,然后递推.
          具体来说,为了避免重复,选择一个固定点x,不管怎样,我们先选择x这个点,然后再从其他剩余的n-1个点中
          分别选0~n-2个点与x构成连通图.然后其他剩余的n-1~1个点他们之间的连接方法随意,也就是2^m种.
          m=完全图的边数=点数*(点数-1)/2.
          初值f[1]=1最后的公式也就是:
          f[n]=2^(n*(n-1)/2)-(f[1]*c[n-1][0]*2^((n-1)*(n-2)/2)+...+f[i]*c[n-1][i-1]*2^((n-i)*(n-i-1)/2+...+f[n-1]*c[n-1][n-2]*2^0)
*/
import java.math.*;  
import java.util.*;   
public class Main {   
      
    public static void main(String[] args) {  
        BigInteger two[] = new BigInteger [2556];  
        two[0] = BigInteger.ONE;  
        for(int i = 1; i <= 2550; i++)  
            two[i] = two[i - 1].multiply(BigInteger.valueOf(2));  
        BigInteger h[] = new BigInteger [55];  
        for(int i = 1; i <= 50; i++){  
            h[i] =two[(i*(i-1))/2];  
        }  
        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]);  
        }  
    }    
}