九度OJ 1044：Pre-Post（先序后序） （n叉树、递归）
- 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
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
****************************************************************/
