uva10247 - Complete Tree Labeling

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本文详细解析了如何通过逆向思维确定深度为d的满k叉树的编号方法数量，深入理解满k叉树的特性及节点编号规律。

Problem B

Complete Tree Labeling!

Input: standard input

Output: standard output

Time Limit: 45 seconds

Memory Limit: 32 MB

A complete k-ary tree is a k-ary tree in which all leaves have same depth and all internal nodes have degreek. Thisk is also known as the branching factor of a tree. It is very easy to determine the number of nodes of such a tree. Given the depth and branching factor of such a tree, you will have to determine in how many different ways you can number the nodes of the tree so that the label of each node is less that that of its descendants. You should assume that for numbering a tree withN nodes you have the(1, 2, 3, N-1, N) labels available.

Input

The input file will contain several lines of input. Each line will contain two integersk andd. Here k is the branching factor of the completek-ary tree andd is the depth of the complete k-ary tree (k>0, d>0, k*d<=21).

Output

For each line of input, produce one line of output containing a round number, which is the number of ways thek-ary tree can be labeled, maintaining the constraints described above.

Sample Input:

2 2

10 1

Sample Output:

3628800

给一个满k叉数从1到N编号，父亲要小于儿子的编号，问深度为d的满k叉数有多少种编号方法。

这道题要有逆向思维，用cnt[i][j]表示深度为j的i叉数有多少个结点，dp[i][j]表示深度为j的i叉树可以编号的方法数。一个深度为j的i叉树是由一个根和i个深度为j-1的i叉树组成的，根是一定要编号为1的。那么根下的第1个深度为j-1的i叉树的编号方法数：从cnt[i][j]-1里选cnt[i-1][j]个编号，再乘以dp[i-1][j]（因为选的这cnt[i-1][j]个编号互不相同，和编号1-cnt[i-1][j]能构成的深度为j-1的i叉树的排列数相同）。同理根下的第2个深度为j-1的i叉树的编号方法数：从刚挑完剩下的cnt[i][j]-1-cnt[i-1][j]中选cnt[i-1][j]个编号，再乘以dp[i-1][j]。

设s=cnt[i][j]-1,m=cnt[i-1][j]，因此dp[i][j]=C(s,m)*dp[i-1][j]*C(s-m,m)*dp[i-1][j]*C(s-2m,m)*dp[i-1][j].....*C(m,m)*dp[i-1][j]

cnt[i][j]=cnt[i][j-1]*2+1

初始化cnt[i][0]=1，dp[i][0]=1。

cnt不用大数，dp要大数。。写个JAVA真恶心。。方法要用static。。

import java.util.*;
import java.math.*;
public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		BigInteger [][]dp=new BigInteger[25][25];
		int [][]cnt=new int[25][25];
		int i,j,k;
		for(i=1;i<=21;i++){
			cnt[i][0]=1;
			dp[i][0]=BigInteger.ONE;
		}
		for(i=1;i<=21;i++)
			for(j=1;j<=21;j++) cnt[i][j]=cnt[i][j-1]*i+1;
		for(i=1;i<=21;i++)
			for(j=1;j<=21/i;j++){
				dp[i][j]=BigInteger.ONE;
				int m=cnt[i][j-1],s=cnt[i][j]-1;
				for(k=s;k>=m;k=k-m){
					dp[i][j]=dp[i][j].multiply(C(k,m)).multiply(dp[i][j-1]);
				}
			}
		Scanner read=new Scanner(System.in);
		while(read.hasNext()){
			int K=read.nextInt();
			int D=read.nextInt();
			System.out.println(dp[K][D]);
		}
	}
	static BigInteger C(int a,int b){
		BigInteger ret=BigInteger.ONE;
		int i;
		for(i=a;i>a-b;i--) ret=ret.multiply(BigInteger.valueOf(i));
		for(i=b;i>0;i--) ret=ret.divide(BigInteger.valueOf(i));
		return ret;
	}
}