hdu1294 Rooted Trees Problem(重复组合+整数拆分+DFS)

最新推荐文章于 2020-01-10 13:44:20 发布

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本文介绍了一种计算特定数量顶点的有根树数量的方法。通过递归地拆分整数并应用组合数学原理来解决该问题。利用Java实现算法，并通过具体实例解释了计算过程。

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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

题意：给出n个结点，问构成一个有根树，有多少种表示形式

思路：用f(n)表示所要求的解，n个结点时，除去一个根结点。先从整数拆分上考虑，实际上是n-1个结点的整数拆分形式，在拆分后，当中会有重复的情况，这又涉及到重复组合的问题。

例如n=5时，4=1+1+1+1=1+1+2=1+3=2+2=4,用加法原理有f(5) = C(f(1)+4-1,4) + C(f(1) + 2 - 1, 2) * C(f(2) + 1 - 1, 1) + C(f(1) + 1 - 1, 1) * C(f(3) + 1 - 1, 1) + C(f(2) + 2 - 1, 2) + C(f(4) + 1 - 1, 1)

注意，输出结果要有long表示，刚开始用int提交出错

代码如下

import java.io.FileReader;
import java.io.OutputStreamWriter;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.PrintWriter;
import java.util.Scanner;

public class Main implements Runnable
{  
    private static final boolean DEBUG = false;
    private Scanner cin;
    private PrintWriter cout;
    private static final int N = 41;
    private static final int[] step = new int[N];
    private static final long[] dp = new long[N];
    private int n;
    
    private void init()
    {
        try {
            if (DEBUG) {
                cin = new Scanner(new BufferedReader(new FileReader("f:\\OJ\\uva_in.txt")));
            } else {
                cin = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
            }
            
            
            cout = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
           
               
            dp[1] = dp[2] = 1;
            for (int i = 3; i < N; i++) {
                dp[i] = 0;
                dfs(0, 1, i - 1, 0);
            }
        } catch (Exception e) {
            e.printStackTrace();
        }  
    }
    
    
    private boolean input()
    {
        if (!cin.hasNextInt()) return false;
        
        n = cin.nextInt();
        return true;
    }
    
    private long c(long n, long m)
    {
        m =  Math.min(m, n - m);
        long ans = 1;
        for (int i = 0; i < m; i++) {
            ans = ans * (n - i) / (i + 1);
        }
        
        return ans;
    }
    
    private void dfs(int dep, int start, int num, int sum)
    {
        if (sum > num) return;
        if (sum == num) {
            long s = 1;
            long k = 1;
            for (int i = 1; i < dep; i++) {
                if (step[i] != step[i - 1]) {
                    s *= c(dp[step[i - 1]] + k - 1, k);
                    k = 0;
                }
                k++;
            }
            
            s *= c(dp[step[dep - 1]] + k - 1, k);
            dp[num + 1] += s;
            return;
        }
        
        for (int i = start; i <= num; i++) {
            step[dep] = i;
            dfs(dep + 1, i, num, sum + i);
        }
    }
    
    private void solve()
    {
       cout.println(dp[n]);
       cout.flush();
    }
    
    @Override
    public void run()
    {
        init();
    
        while (input()) 
        {
            solve();
        }
    }
    
    public static void main(String[] args) 
    {
        // TODO code application logic here
       new Thread(new Main()).start();
    }
}