catalan数 + 二叉树 —— POJ 1095

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Trees Made to Order

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7008		Accepted: 4000

Description

We can number binary trees using the following scheme:

The empty tree is numbered 0.

The single-node tree is numbered 1.

All binary trees having m nodes have numbers less than all those having m+1 nodes.

Any binary tree having m nodes with left and right subtrees L and R is numbered n such that all trees having m nodes numbered > n have either Left subtrees numbered higher than L, or A left subtree = L and a right subtree numbered higher than R.

The first 10 binary trees and tree number 20 in this sequence are shown below:



Your job for this problem is to output a binary tree when given its order number.

Input

Input consists of multiple problem instances. Each instance consists of a single integer n, where 1 <= n <= 500,000,000. A value of n = 0 terminates input. (Note that this means you will never have to output the empty tree.)
Output

For each problem instance, you should output one line containing the tree corresponding to the order number for that instance. To print out the tree, use the following scheme:

A tree with no children should be output as X.

A tree with left and right subtrees L and R should be output as (L')X(R'), where L' and R' are the representations of L and R.

If L is empty, just output X(R').

If R is empty, just output (L')X.

Sample Input

1
20
31117532
0

Sample Output

X
((X)X(X))X
(X(X(((X(X))X(X))X(X))))X(((X((X)X((X)X)))X)X)

题意：

我们可以用数字来表示唯一的二叉树：

1）空树用编号 0 表示；

2）只有一个结点的树用编号 1 表示；

3）所有含有 m 个结点的二叉树的编号小于含有 m + 1 个结点的二叉树的编号；

4）假如一颗二叉树 T1 有 m 个结点，编号为 n ，左子树的编号为 L，右子树的编号为 R；则要使跟 T1 有相同结点数（即 m ）的二叉树 T2 的编号大于 n ，需要满足这两个条件的任意一个：1、T2 的左子树编号大于 L ；2、T2 的左子树编号等于 L，但 T2 的右子树编号大于 R；

思路：

设 f(n) = 有 n 个结点的二叉树一共有多少种表示方法(编号)，由乘法原理易知：

f(0) = 1;

f(1) = 1;

f(2) = 2;

f(3) = f(0)*f(2) + f(1)*f(1) + f(2)*f(0) = 5;

f(4) = f(0)*f(3) + f(1)*f(2) + f(2)*f(1)
+ f(3)*f(0) = 14;

...

所以~这是 catalan 数，可以根据递推公式 f(n) = f(n - 1) * ((4*n - 2) / (n + 1)) 计算出来，记为catalan[]数组；对于一个请求 n :

结点数 m = { j | catalan[0] + catalan[1] + ... + catalan[j] >= n }

在 m 个结点的所有二叉树中编号的排位 pos = n - catalan[0] + catalan[1] + ... + catalan[m-1];

接着是构建二叉树，用递归的思路，我们可以用 BuildTree(m, pos) 来表示构建一颗含有 m 个结点，在 m 个结点的所有二叉树的编号中排位为 pos 的一颗二叉树。如果左子树的结点个数为 i ，则右子树的结点个数为 m - i - 1；我们知道 m个结点的二叉树一共有 f(m) = f(0)*f(m -
1) + f(1)*f(m - 2) + f(2)*f(m - 3) + f(m - 1)*f(0) 个不同的编号；我们需要知道 pos 是在哪个 f(i)*f(m - i - 1) 里面，从而确定左右子树的结点个数进行递归；设 sf(j) = f(0)*f(1) + f(1)*f(m
- 2) + ... + f(j)f(m - j - 1) ，则 i = {j | sf(j-1) < pos <= sf(j)}。

用 next_pos 表示 在 f(i)*f(m - i - 1) 种编号中的排位，则 next_pos = pos - sf(i-1)；那左子树跟右子树的排位分别是多少呢？可以用类似于进位的思想去想。比如左子树有
2 个结点， 右子树有 3 个结点，则它们一共有 f(2)*f(3) = 10 个编号，next_pos 肯定是 1 ~ 10 的某个数。根据题目意思，它是按右中左的顺序编号的；则我们可以用 next_pos / f(3) 来表示左子树在 f(2) 个编号中的排位 left_pos；用 next_pos % f(3) 来表示右子树在 f(3) 个编号中的排位 right_pos；由于排位要从 1 开始，所以应写成 left_pos = ((next_pos
- 1) / f(m - i - 1)) + 1; right_pos = ((next_pos - 1) % f(m - i - 1)) + 1; 最后进行递归调用 BuildTree(i, left_pos) 和 BuildTree(m - i - 1, right_pos) 就可以了。


在递归的过程输出结果就不需要建树。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int catalan[20] = {
	1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700
};
int sum[20] = {
	0, 1, 3, 8, 22, 64, 196, 625, 2055, 6917, 23713, 82499, 290511, 1033411, 3707851, 13402696, 48760366, 178405156, 656043856
};

void BuildTree(int m, int pos)
{
	int cnt = 0;
	int i, s = 0;
	if(0 == m) return;
	/*
	if(1 == m){
		putchar('X');
		return;
	}
	*/
	for(i = 0; i < m; i++){
		int tmp = catalan[i]*catalan[m-i-1];
		if(cnt < pos && pos <= cnt+tmp){
			s = pos - cnt;
			break;
		}
		cnt += tmp;
	}
	if(i){
		putchar('(');
		BuildTree(i, (s-1)/catalan[m-i-1] + 1);
		putchar(')');
	}
	putchar('X');
	if(m - i - 1){
		putchar('(');
		BuildTree(m - i - 1, (s-1)%catalan[m-i-1]+1);
		putchar(')');
	}
}

int main()
{
#if 0
	freopen("in.txt","r",stdin);
#endif
	int n, m, pos;
	while(scanf("%d", &n), n){
		int i;
		for(i = 1; i <= 18; i++)
			if(sum[i] >= n){
				m = i;
				pos = n - sum[m-1];
				break;
			}
		BuildTree(m, pos);
		putchar('\n');
	}
	return 0;
}