考研编程练习----m叉树先序和后序所包含的情况

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

经典代码：

 

#include <stdio.h>

#include <string.h>

char pre[30], post[30];

int m;

int find(char* p, char x) {

    int i=0;

    while (p[i]!=x) i++;

    return i;

}

typedef long long ll;

ll C(int a, int b) {

    ll u = 1;

    ll d = 1;

    while (b) {

    u *= a--;

    d *= b--;

    }

    return u/d;

}

ll test(char* p, char* q, int n) {

    if (n==0) return 1;

    ll f = 1;

    int c = 0;

    int i;

    while (n) {

    c++;

    i = find(q,*p);

    f *= test(p+1,q,i);

    p += i+1;

    q += i+1;

    n -= i+1;

    }

    return f * C(m,c);

}

int main() {

    while(scanf("%d%s%s",&m,pre,post)==3) {

    printf("%lld\n",test(pre+1,post,strlen(pre)-1));

    }  

    return 0;

}

 

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