九度OJ 1044：Pre-Post（先序后序） （n叉树、递归）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：701

解决：398

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

思路：

求对于m叉树而言其每层的叶子节点的组合方式有多少种。

由先序和后序序列其实可以却确定每一层的叶子节点的个数，以及哪些是这一层的叶子节点，唯一不确定的就是这些节点的位置（但是由先序可以确定这些叶子节点相对位置是确定的），比如第i层有n个叶子节点（由先序和后序结合判定出），那么这层就有c(n,m)种组合方式，然后确定某个叶子节点的子树，对其进行递归求解。

代码：

#include <stdio.h>
#include <string.h>
 
#define M 20
#define N 26
 
int m;
 
long long C(int m, int k)
{
    int i;
    long long c = 1;
    for (i=m; i>k; i--)
        c *= i;
    for (i=m-k; i>0; i--)
        c /= i;
    return c;
}
 
long long prepost(char s1[], char s2[])
{
    int len = strlen(s1);
    if (len == 0 || len == 1)
        return 1;
 
    char root;
    int c;
    char sch1[M][N+1], sch2[M][N+1];
    int i, j, k;
    c = 0;
    i = 1;
    while (i < len)
    {
        root = s1[i];
        j = i-1;
        k = 0;
        do
        {
            sch1[c][k] = s1[i];
            sch2[c][k] = s2[j];
            k++;
            i++;
        } while (s2[j++] != root);
        sch1[c][k] = '\0';
        sch2[c][k] = '\0';
        c++;
    }
 
    long long count = C(m, c);
    for (i=0; i<c; i++)
        count *= prepost(sch1[i], sch2[i]);
    return count;
}
 
int main(void)
{
    char s1[N+1], s2[N+1];
 
    while (scanf("%d", &m) != EOF)
    {
        scanf("%s%s", s1, s2);
        //printf("%lld\n", C(6, 2));
        printf("%lld\n", prepost(s1, s2));
    }
 
    return 0;
}
/**************************************************************
    Problem: 1044
    User: liangrx06
    Language: C
    Result: Accepted
    Time:0 ms
    Memory:912 kb
****************************************************************/