POJ1737 Connected Graph

题意

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Connected Graph

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3865		Accepted: 1812

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

给定n个点，每两个点间至多连一条无向边，问任意连边，这些点能够成的无向连通图的个数

分析

参照ustcscuwf的题解。

典型的递推，裸算几乎不可能，生成树图的个数都要通过母函数进行推导和计算，这个更加复杂，递推公式为F(n)=sigma(F(n-k)F(k)C(n-2,k-1)*(2^k-1))。显然要用大数计算，最后的数字有三百多位。

这个递推还是比较好理解的，就是分成两个子连通部分，要想整个连通，那么两个子连通图中间必然要有一条边连接，在划分子连通时，注意固定子连通中的各一个点，防止重复计数。规定连边必须向固定的点（i）连边。

时间复杂度\(O(n^2 S^2)\)

代码

#include<iostream>
#include<cstring>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
    while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;

co int N=60,S=600;
int n;
struct A{
	int a[S],len;
	il A operator/(co int x)co{
		A ans;
		memset(ans.a,0,sizeof ans.a);
		ans.len=0;
		int num=0;
		for(int i=len;i;--i){
			num=num*10+a[i];
			ans.a[i]=num/x;
			num%=x;
			if(!ans.len&&ans.a[i]) ans.len=i;
		}
		return ans;
	}
	il A operator+(co A&x)co{
		A ans;
		memset(ans.a,0,sizeof ans.a);
		for(int i=1;i<=max(len,x.len);++i){
			ans.a[i]+=a[i]+x.a[i];
			ans.a[i+1]=ans.a[i]/10;
			ans.a[i]%=10;
		}
		ans.len=max(len,x.len);
		if(ans.a[ans.len+1]) ++ans.len;
		return ans;
	}
	il A operator*(co A&x)co{
		A ans;
		memset(ans.a,0,sizeof ans.a);
		for(int i=1;i<=len;++i)
			for(int j=1;j<=x.len;++j){
				ans.a[i+j-1]+=a[i]*x.a[j];
				ans.a[i+j]+=ans.a[i+j-1]/10;
				ans.a[i+j-1]%=10;
			}
		ans.len=len+x.len-1;
		if(ans.a[ans.len+1]) ++ans.len;
		return ans;
	}
}f[N],p[N];
il A C(int x,int y){
	A ans;
	ans.len=ans.a[1]=1;
	for(int i=y,j=1;j<=x;--i,++j){
		int t=i;
		A tmp;
		tmp.len=0;
		while(t){
			tmp.a[++tmp.len]=t%10;
			t/=10;
		}
		ans=ans*tmp/j;
	}
	return ans;
}
il void print(co A&x){
	for(int i=x.len;i;--i) printf("%d",x.a[i]);
	puts("");
}
int main(){
	for(int i=1;i<=50;++i){
		ll t=(1LL<<i)-1;
		while(t){
			p[i].a[++p[i].len]=t%10;
			t/=10;
		}
	}
	f[1].len=f[2].len=f[1].a[1]=f[2].a[1]=1;
	for(int i=3;i<=50;++i)
		for(int j=1;j<=i-1;++j)
			f[i]=f[i]+C(j-1,i-2)*f[j]*f[i-j]*p[j];
	while(read(n)) print(f[n]);
	return 0;
}