POJ 1737 Connected Graph

题意：求 n 个点的无向简单带标号联通图个数。

记 f[n] 是所求答案，g[n] 是 n 个点的不联通图个数，f[n] + g[n] = h[n] = 2 ^ ((n*(n-1)/2)

设 1 所在联通块有 k 个点， 1<=k<n

Connected Graph

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 2945 Accepted: 1445

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

12340

Sample Output

11438

Source

LouTiancheng@POJ

/** * Created by OI_lover on 2015/4/9. */import java.io.*;import java.util.*;import java.math.BigInteger;import static java.math.BigInteger.*;public class Main {    public static final int N = 55;    public static void main(String[] args) throws Exception    {        BigInteger []h = new BigInteger[N];        BigInteger []f = new BigInteger[N];        BigInteger []g = new BigInteger[N];        BigInteger p2[] = new BigInteger[N*N];        BigInteger [][]C = new BigInteger[N][N];        for (int i=0;i<N;i++)        {            for (int j=0;j<=i;j++)            {                if (i==j || j==0) C[i][j] = valueOf(1);                else C[i][j] = C[i-1][j-1].add(C[i-1][j]);            }        }        p2[0] = ONE;        for (int i=1;i<N*N;i++)            p2[i] = p2[i-1].multiply(valueOf(2));        for (int i=0;i<N;i++)            h[i] = p2[i*(i-1)/2];        f[1] = BigInteger.valueOf(1);        g[1] = BigInteger.valueOf(0);        for (int n=2;n<N;n++)        {            g[n] = BigInteger.ZERO;            for (int k = 1;k < n; k++)            {                BigInteger t = C[n-1][k-1].multiply(f[k]);                t = t.multiply(h[n-k]);                g[n] = g[n].add(t);                f[n] = h[n].subtract(g[n]);            }        }        Scanner in = new Scanner(System.in);        while (in.hasNext())        {            int n = in.nextInt();            if (n == 0) break;            System.out.println(f[n]);        }    }}