九度oj 1044

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 cba
2 abc bca
10 abc bca
13 abejkcfghid jkebfghicda

样例输出：

4
1
45
207352860

来源：
2008年上海交通大学计算机研究生机试真题

#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++; //num代表串被分成了几段(例如 (bejkcfghid,jkebfghicd) = (bejk, cfghi, d) )

k = i + 1;

break;

}

}

//cout << pre << " " << post <<" " << t1 << " =" << num << endl << endl;

sum *= c[num]
; //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;

else

c[j][i] = c[j - 1][i - 1] + c[j][i - 1];

}

}

}

int main() {

string pre, post;

getsc();

while ((cin >> n >> pre >> post) && n) {

cout << test(pre, post) << endl; //printf("%ld\n",test(pre,post));

}

return 0;

}