九度 1044：Pre-Post 递归求n叉树结构个数

题目描述：

        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:

    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 cba2 abc bca10 abc bca13 abejkcfghid jkebfghicda

样例输出：

4145207352860

来源：

2008年上海交通大学计算机研究生机试真题

题目大意：对于n叉树，给出先序遍历和后续遍历，求可能的个数。

例如：

10 abc bca

根节点为a是确定的，接下来是 bc bc

可知b,c在同一级别，有C（10,2）=45 (10个位置中取两个)

2 abc cba

同样根节点为a，然后是 bc cb

b,c在两层 C（2,1） * C（2,1）=4

对于这个就有些复杂了,

13 abejkcfghid jkebfghicda

第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3)

再继续递归下去，直到字符串长度为1

#include <stdio.h>#include <string>#include <iostream>using namespace std;int c[21][21];int n;long long test(string pre, string post) {long long sum = 1;int num = 0; int k = 0, i;pre.erase(pre.begin());post=post.substr(0, post.length()-1);while (k < pre.length()) {for (i = 0; i < post.length(); i++)if (pre[k] == post[i]) {sum *= test(pre.substr(k, i - k + 1),post.substr(k, i - k + 1));//num代表串被分成了几段(例如 (bejkcfghid,jkebfghicd) = (bejk, cfghi, d) , num=3)num++;k = i + 1;break;}}//cout << pre << "  " << post <<"  " << t1 << " =" << num << endl << endl;sum *= c[num][n]; //从n中取num个的取法个数return sum;}void getsc() {int i, j;c[0][1] = c[1][1] = 1;for (i = 2; i < 21; i++) {c[0][i] = 1;for (j = 1; j <= i; j++){if (i == j)c[j][i] = 1;elsec[j][i] = c[j - 1][i - 1] + c[j][i - 1];}}}int main() {freopen("in.txt", "r", stdin);string pre, post;getsc();while ((cin >> n >> pre >> post) && n) {cout << test(pre, post) << endl;}return 0;}