POJ 1737: Connected Graph 组合数学 高精度

Connected Graph

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3459		Accepted: 1679

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

啊啊啊 太棒了 这个题的英文我都认识

设 f[i] 表示 i 个点的答案

之后考虑加入 n 号点 统计即可以 n 为枢纽用不同的方案链接所有的点

套路枚举 1 所在联通块就好

更详细可以看这个

http://blog.youkuaiyun.com/blackjack_/article/details/78377228

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<bitset>
#include<queue>
#include<set>
#include<map>
using namespace std;

inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
void print(int x)
{if(x<0)putchar('-'),x=-x;if(x>=10)print(x/10);putchar(x%10+'0');}

const int N=210,rad=1000;

struct bint
{
	int a[N];
	bint(){memset(a,0,sizeof(a));a[0]=1;}
	
	friend bint operator +(const bint &x,const bint &y)
	{
		bint res;
		res.a[0]=max(x.a[0],y.a[0]);
		for(int i=1;i<=res.a[0];++i)
			res.a[i]=x.a[i]+y.a[i];
		for(int i=1;i<=res.a[0];++i)
		{
			if(res.a[i]>=rad)
			{
				res.a[i+1]+=res.a[i]/rad;res.a[i]%=rad;
				res.a[0]=max(res.a[0],i+1);
			}
		}
		return res;
	}
	
	friend bint operator *(const bint &x,int y)
	{
		bint res=x;
		for(int i=1;i<=res.a[0];++i) res.a[i]*=y;
		for(int i=1;i<=res.a[0];++i)
		{
			if(res.a[i]>=rad)
			{
				res.a[i+1]+=res.a[i]/rad;res.a[i]%=rad;
				res.a[0]=max(res.a[0],i+1);
			}
		}
		return res;
	}
	
	friend bint operator *(const bint &x,const bint &y)
	{
		bint res;res.a[0]=x.a[0]+y.a[0]-1;
		for(int i=1;i<=x.a[0];++i)
			for(int j=1;j<=y.a[0];++j)
				res.a[i+j-1]+=x.a[i]*y.a[j];
		for(int i=1;i<=res.a[0];++i)
		{
			if(res.a[i]>=rad)
			{
				res.a[i+1]+=res.a[i]/rad;res.a[i]%=rad;
				res.a[0]=max(res.a[0],i+1);
			}
		}
		return res;
	}
	
	friend void print_bint(const bint &x)
	{
		printf("%d",x.a[x.a[0]]);
		for(int i=x.a[0]-1;i;i--)
			printf("%03d",x.a[i]);
		putchar('\n');
	}
	
}c[51][51],f[51],pw[51];

void initial()
{
	register int i,j,s;
	bint tmp;tmp.a[1]=1;
	for(i=1;i<=50;++i) pw[i]=pw[i-1]*2+tmp;
	c[0][0].a[1]=1;
	for(i=1;i<=50;++i)
	{
		c[i][0].a[1]=1;
		for(j=1;j<=i;++j)
			c[i][j]=c[i-1][j]+c[i-1][j-1];
	}
	f[1].a[1]=1;
	for(i=2;i<=50;++i)
		for(s=1;s<=i-1;++s)
			f[i]=f[i]+c[i-2][s-1]*f[s]*f[i-s]*pw[s];
}

int main()
{
	initial();
	while(1)
	{
		int n=read();
		if(!n) break;
		print_bint(f[n]);
	}
	return 0;
}