hdu 1234 Rooted Trees Problem 有重复的组合+乘法原理-优快云博客

本文链接：https://blog.youkuaiyun.com/jyysc2010/article/details/10211967

本文介绍了一种计算特定数量顶点的根树数目方法。通过定义树与根树的概念，利用递归算法与组合数学原理，实现了对于任意n值（1≤n≤40）的根树数量Tn的有效计算。

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Rooted Trees Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 443 Accepted Submission(s): 147

Problem Description

Give you two definitions tree and rooted tree. An undirected connected graph without cycles is called a tree. A tree is called rooted if it has a distinguished vertex r called the root. Your task is to make a program to calculate the number of rooted trees with n vertices denoted as Tn. The case n=5 is shown in Fig. 1.

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer means n(1<=n<=40) .

Output

For each test case, there is only one integer means Tn.

Sample Input

Sample Output

Author

SmallBeer(CML)

Source

杭电ACM集训队训练赛（VIII）

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参考： http://www.cppblog.com/Yuan/archive/2010/10/22/130929.html

#include<cstdio>
#include<iostream>
using namespace std;
long long t[50];
int cnt[50] , n;

long long gcd(long long a, long long b)
{
    return b?gcd(b,a%b):a;
}

long long C(long long n, long long m)
{
    if(m > n - m ) m = n - m;
    long long mul = 1 , div = 1;
    for(int i = 0; i < m; i++)
    {
        mul *= (n-i);
        div *= (i+1);
        long long g = gcd(mul,div);
        mul/=g , div /= g;
    }
    return mul / div;
}

void dfs(int start, int left)
{
    if( left == 0 )
    {
        long long ans = 1;
        for(int i = 1; i<=n ; i ++)
        {
            if(cnt[i] == 0)continue;
            ans = ans * C(t[i] - 1 + cnt[i] , cnt[i]);
        }
        t[n] += ans;
        return ;
    }
    if(start > left )return;
    cnt[start]++;
    dfs(start , left - start);
    cnt[start]--;
    dfs(start+1 , left);
}

int main()
{

    t[1] = t[2] = 1;
    for(n = 3; n <= 40 ; n++)
    {
        dfs(1,n-1);
    }
    while( ~scanf("%d",&n) )
        printf("%I64d\n",t[n]);
    return 0;
}