poj1240 Pre-Post-erous!

照例先上题目：

1:Pre-Post-erous!
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总时间限制: 1000ms 内存限制: 65536kB
描述
We are all familiar with pre-order, in-order and post-order traversals of binary trees. A common problem in data structure classes is to find the pre-order traversal of a binary tree when given the in-order and post-order traversals. Alternatively, you can find the post-order traversal when given the in-order and pre-order. However, in general you cannot determine the in-order traversal of a tree when given its pre-order and post-order traversals. Consider the four binary trees below:

a a a a
/ / \ \
b b b b
/ \ / \
c c c c

All of these trees have the same pre-order and post-order traversals. This phenomenon is not restricted to binary trees, but holds for general m-ary trees as well.
输入
Input will consist of multiple problem instances. Each instance will consist of a line of the form
m s1 s2
indicating that the trees are m-ary trees, s1 is the pre-order traversal and s2 is the post-order traversal.All traversal strings will consist of lowercase alphabetic characters. For all input instances, 1 <= m <= 20 and the length of s1 and s2 will be between 1 and 26 inclusive. If the length of s1 is k (which is the same as the length of s2, of course), the first k letters of the alphabet will be used in the strings. An input line of 0 will terminate the input.
输出
For each problem instance, you should output one line containing the number of possible trees which would result in the pre-order and post-order traversals for the instance. All output values will be within the range of a 32-bit signed integer. For each problem instance, you are guaranteed that there is at least one tree with the given pre-order and post-order traversals.
样例输入
2 abc cba
2 abc bca
10 abc bca
13 abejkcfghid jkebfghicda
0
样例输出
4
1
45
207352860
来源
East Central North America 2002

1. 在不作其他规定的情况下，对于一般的m叉树，dfs的方式仅有先根遍历和后根遍历。先根遍历和后根遍历都是递归定义的过程。
2. 先根遍历：访问根结点；然后按顺序先根遍历第0、1、2...m-1棵子树。后根遍历：按顺序后根遍历第0、1、2...m-1棵子树；然后访问根结点。

若给定m叉树的先根遍历和后根遍历序列，我们可以重建树的大致形状。举个例子，样例输入中的"13 abejkcfghid jkebfghicda":

最后得：

因此共计C(13, 3)*C(13, 1)*C(13, 4)*C(13, 2)=207352860种可能性。

代码清单：

#include 
#include 
#include 
#include 
using namespace std;

#include 

#define MAXLEN 30
char pre[MAXLEN];
char post[MAXLEN];

vector branch_vec;

//计算n!
int factorial(int n)
{
	int result=1;

	if (n==0) return 1;
	else
	{
		for (int i=1; i<=n; ++i) result *= i;
		return result;
	}
}

//计算C(n, m)
int calc_C(int n, int m)
{
	unsigned int tmp=1;
	for (int i=0; i