C++——【USACO 3.4.1】——American Heritage

American Heritage

Farmer John takes the heritage of his cows very seriously. He is not, however, a truly fine bookkeeper. He keeps his cow genealogies as binary trees and, instead of writing them in graphic form, he records them
in the more linear `tree in-order' and `tree pre-order' notations.
Your job is to create the `tree post-order' notation of a cow's heritage after being given the in-order and pre-order notations. Each cow name is encoded as a unique letter. (You may already know that you can frequently
reconstruct a tree from any two of the ordered traversals.) Obviously, the trees will have no more than 26 nodes.
Here is a graphical representation of the tree used in the sample input and output:

C
/   \
/     \
B       G
/ \     /
A   D   H
/ \
E   F

The in-order traversal of this tree prints the left sub-tree, the root, and the right sub-tree.
The pre-order traversal of this tree prints the root, the left sub-tree, and the right sub-tree.
The post-order traversal of this tree print the left sub-tree, the right sub-tree, and the root.

PROGRAM NAME: heritage

INPUT FORMAT

Line 1:	The in-order representation of a tree.
Line 2:	The pre-order representation of that same tree.

SAMPLE INPUT (file heritage.in)

ABEDFCHG
CBADEFGH

OUTPUT FORMAT

A single line with the post-order representation of the tree.

SAMPLE OUTPUT (file heritage.out)

AEFDBHGC

农夫约翰非常认真地对待他的奶牛们的血统。然而他不是一个真正优秀的记帐员。他把他的奶牛们的家谱作成二叉树，并且把二叉树以更线性的“树的中序遍历”和“树的前序遍历”的符号加以记录而不是用图形的方法。

你的任务是在被给予奶牛家谱的“树中序遍历”和“树前序遍历”的符号后，创建奶牛家谱的“树的后序遍历”的符号。每一头奶牛的姓名被译为一个唯一的字母。（你可能已经知道你可以在知道树的两种遍历以后可以经常地重建这棵树。）显然，这里的树不会有多于26个的顶点。 这是在样例输入和样例输出中的树的图形表达方式：

C
/ \
/   \
B     G
/ \   /
A   D H
/ \
E   F

树的中序遍历是按照左子树，根，右子树的顺序访问节点。

树的前序遍历是按照根，左子树，右子树的顺序访问节点。

树的后序遍历是按照左子树，右子树，根的顺序访问节点。

格式

PROGRAM NAME: heritage

INPUT FORMAT:

(file heritage.in)

第一行： 树的中序遍历

第二行： 同样的树的前序遍历

OUTPUT FORMAT:

(file heritage.out)

单独的一行表示该树的后序遍历。

SAMPLE INPUT

ABEDFCHG
CBADEFGH

SAMPLE OUTPUT

AEFDBHGC

/*
ID : mcdonne1
LANG : C++
TASK : heritage
*/
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
struct node *left;
struct node *right;
char c;
};
void hxbl(char *s1,char *s2,int len)
{
if(len==0) return;
node *n=new node;
n->c=*s2;
int index;
for(index=0;index<len;++index)
if(s1[index]==*s2) break;
hxbl(s1,s2+1,index);
hxbl(s1+index+1,s2+index+1,len-index-1);
cout<<n->c;
}
int main()
{
freopen("heritage.in","r",stdin);
freopen("heritage.out","w",stdout);
char s1[30],s2[30];
scanf("%s%s",s1,s2);
hxbl(s1,s2,strlen(s1));
putchar('\n');
return 0;
}