USACO3.4.2American 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 
题解：《算法竞赛入门经典》的106页的《二叉树重建》和此题一模一样，都是已知前序遍历和中序遍历，求后序遍历。因为前序遍历的第一个元素就是二叉树的根，因此只需在中序遍历中找到它，我们就可以知道左子树和右子树的前序遍历和后序遍历了，可以用递归来实现。
直接上书上的代码。。。。

/*
ID:spcjv51
PROG:heritage
LANG:C
*/
#include<stdio.h>
#include<string.h>
#define MAXN 30
char s1[MAXN],s2[MAXN],ans[MAXN];
void build(int n,char *s1,char *s2,char *s)
{   int p;
    if(n<=0) return ;
    p=strchr(s2,s1[0])-s2;
    build(p,s1+1,s2,s);
    build(n-p-1,s1+p+1,s2+p+1,s+p);
    s[n-1]=s1[0];
}
int main(void)
{
    freopen("heritage.in","r",stdin);
    freopen("heritage.out","w",stdout);
    int n;
    scanf("%s%s",s1,s2);
    n=strlen(s1);
    build(n,s2,s1,ans);
    ans[n]='\0';
    printf("%s\n",ans);
    return 0;
}