POJ1240 m叉树

Pre-Post-erous!

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1915		Accepted: 1158

Description

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:

    a   a   a   a
    /   /     \   \
   b   b       b   b
  /     \     /     \
 c       c   c       c

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

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.

Output

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.

Sample Input

2 abc cba
2 abc bca
10 abc bca
13 abejkcfghid jkebfghicda
0

Sample Output

4
1
45
207352860
题意：已知前序遍历和后序遍历，求中序遍历的种数。（如果已知先序和中序，那么后序唯一，如果已知中序和后序，先序也是唯一的）
思路：。。。。。。

#include<stdio.h>
#include<string.h>
#include <stdio.h>
#include <string.h>
int N, M;
char s1[30],s2[30];
int C(int n, int k)                          //计算组合数C(n, k)
{
    int i,res=1;
    if(k>n/2) k=n-k;
    for(i=1;i<=k;++i)
    {
        res=res*(n-i+1)/i;
    }
    return res;
}
int DP(int m,char *pre,char *post)
{
    int left=1,right=0,nchild=0,res=1;//res: 结果
    //left: pre的指针，指向子树根节点。根节点不用看了，所以从1开始
    //right: post的指针，指向子树根节点
    //nchild: 有多少个子树

    while(left<m)             //枚举每棵子树
    {
        nchild++;                //多一棵子树
        while(pre[left]!=post[right])
        {right++;}
        res*=DP((right+1)-left+1,pre+left,post+left-1);
        //对每棵子树递归进行动态规划，由乘法原理，应该是把结果乘起来
        left=right+2;      //设置寻找下一棵子树
    }
    return res*C(N,nchild);//因为是m叉树
}
int main()
{
    while(scanf("%d",&N),N)
    {
        scanf("%s%s",s1,s2);
        M=strlen(s1);//结点数
        printf("%d\n",DP(M,s1,s2));//递归求解
    }
    return 0;
}