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.
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.
2 abc cba 2 abc bca 10 abc bca 13 abejkcfghid jkebfghicda
4 1 45 207352860
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; else c[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; }