B - Tree Recovery

Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.
This is an example of one of her creations:

                                           D

                                          / \

                                         /   \

                                        B     E

                                       / \     \

                                      /   \     \

                                     A     C     G

                                                /

                                               /

                                              F

To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree). For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).

Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!

Input

The input will contain one or more test cases.
Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)
Input is terminated by end of file.

Output

For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).

Sample Input

DBACEGF ABCDEFG
BCAD CBAD

Sample Output

ACBFGED
CDAB

题意：给你前序和中序，求后序；只要找到规律就可以，前序的第一个一定是子树的根节点，在中序的根节点左边的是根节点的左子树，一直找下去，直到不能找了输出

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+100;
const double eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int inf=0xfffffff;
const int N=1005;
char a[30],b[30];
void find(int l1,int r1,int l2,int r2)
{
	if(l1>r1) return ; 
	int y=l2; 
	while(b[y]!=a[l1]) y++; 
	find(l1+1,r1-(r2-y),l2,y-1);
	find(r1+1-(r2-y),r1,y+1,r2);
	pf("%c",a[l1]);
}
int main()
{
	while(~sf("%s%s",a,b))
	{
		int l=strlen(a);
		find(0,l-1,0,l-1);
		pf("\n");
	}
	return 0;
}