poj Connected Graph

Connected Graph

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2197		Accepted: 1071

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

Source

LouTiancheng@POJ

 

这是传说中的楼教主出的题，果然不一般呀；看的别人的题解呀，用java写的，牛呀

公式是f[n]=f[k]*f[n-k]*c[n-2][k-1]*((2^k)-1) (c[n-2][k]表示组合数,1<=k<n);考虑一个完整的联通图，可以标记两个点1，2。将点1，点2分别划分在两个子联通图中分别 为g1，g2。在g1中最少要有一个点与g2中的点链接。这样的方式共2^k-1中，而g1总有c[n-2][k-1]个。将他们乘在一起就有了上面的式 子(转载)。

import java.io.*;
import java.math.*;
import java.util.*;
public classMain{
 static BigInteger p=BigInteger.valueOf(2);
 public static void main(String[] args)
 {
  Scanner cin=new Scanner(System.in);
     int i,j,k;
     BigInteger c[][]=new BigInteger[52][52];
     BigInteger f[]=new BigInteger [52];
     for(i=0;i<=50;i++)
     c[i][0]=c[i][i]=BigInteger.ONE;
     for(i=1;i<=50;i++)
      for(j=1;j<i;j++)
        c[i][j]=c[i-1][j-1].add(c[i-1][j]);
      
     int n;
     
     f[1]=BigInteger.ONE;
     f[2]=BigInteger.ONE;
     f[3]=BigInteger.valueOf(4);
     for(i=4;i<=50;i++)
     {
      f[i]=BigInteger.ZERO;
      for(k=1;k<i;k++)
      {
       BigInteger t=BigInteger.ZERO;
       t=f[k].multiply(f[i-k]).multiply(c[i-2][k-1]);
       t=t.multiply(p.pow(k).subtract(BigInteger.ONE));
       f[i]=f[i].add(t);
      }
     }
     while(cin.hasNext())
     {
      n=cin.nextInt();
      if(n==0)
      break;
      if(n==1||n==2)
      {
       System.out.println(1);
       continue;
      }
         System.out.println(f[n]);
     }
 }
}