HDU 1294 Rooted Trees Problem

 

 

原来是C(n,m)越界了- -难怪一直不出答案- -#

Rooted Trees Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 472    Accepted Submission(s): 161

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

1

2

5

 

 

Sample Output

1

1

9

 

 

Author

SmallBeer(CML)

 

 

Source

杭电ACM集训队训练赛（VIII）

 

 

Recommend

lcy

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 105
#define mod 1000000007
#define ll long long
#define INF 0x7fffffff
int n, m, s;
ll f[maxn];
int cnt[maxn];
ll gcd(ll a, ll b){return b ? gcd(b, a%b) : a;}
ll C(ll n, ll m)
{
    m = min(m, n - m);
    ll mul = 1, div = 1;
    for (int i = 0; i < m; i++)
    {
        mul *= (n - i);
        div *= (i + 1);
        ll g = gcd(mul, div);
        mul /= g, div /= g;
    }
    return mul / div;
}
void dfs(int n,int a,int step,ll t){
    if (t>n)return;
    if (t == n){
        ll s = 1;
        int k = 1;
        for (int i = 1; i < step; i++){
            if (cnt[i] == 0)continue;
            if (cnt[i] != cnt[i - 1]){
                s *= C(f[cnt[i-1]] + k - 1, k);
                k = 0;
            }
            k++;
        }
        s *= C(f[cnt[step-1]] + k - 1, k);
        f[n+1] += s;
        return;
    }
    for (int i = a; i <= n; i++){
        cnt[step] = i;
        dfs(n, i, step + 1, t + i);
    }
}
int main(){
    int cas = 1;
    int t;
    /*scanf("%d", &t);
    while (t--){
        scanf("%d", &n));
    }*/
    f[1] = f[2] = 1;
    for (int i = 3; i <= 40; i++){ f[i] = 0; dfs(i - 1, 1, 0, 0); }
    while (~scanf("%d", &n)){
        printf("%I64d\n", f[n]);
    }
    return 0;
}