USACO: 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 

 

解题思路：

这道题没什么复杂度上的难度，只要想清楚了解法就是秒杀

用一个数组保存每个字母所在的子树在字符串中的范围，根据前序串逐个扫描，缩小子树的范围，并对中序串作交换，扫描完成后中序串也就交换完毕变成了后序串

 

比较直观地举个例子：

Tree: C / / D E / / G I / H in-order:HGDICE pre-order:CDGHIE Scan the pre-order string: 012345 C D E G H I HGDICE 0~5 0~5 0~5 0~5 0~5 0~5 C HGDIEC / 0~3 4~4 0~3 0~3 0~3 D HGIDEC / / 4~4 0~1 0~1 2~2 G HGIDEC / / 4~4 / 0~0 2~2 H HGIDEC / / 4~4 / / 2~2 I HGIDEC / / 4~4 / / / E HGIDEC / / / / / / post-order: HGIDEC

/* ID: geterns1 PROG: heritage LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; inline char idx(char &c){ return c - 'A'; } int main(){ freopen("heritage.in", "r", stdin); freopen("heritage.out", "w", stdout); const short size = 26; short len; short suba[size], subb[size], map[size]; //range of subtree and a map of string address char inOrder[size + 1], preOrder[size + 1]; //initial data // scanf("%s%s", inOrder, preOrder); //initialization len = strlen(inOrder); for (int i = 0; i < size; i ++){ suba[i] = 0; subb[i] = len - 1; } for (int i = 0; i < len; i ++) map[idx(inOrder[i])] = i; // for (int i = 0; i < len; i ++){ //left subtree for (int j = suba[idx(preOrder[i])]; j < map[idx(preOrder[i])]; j ++) subb[idx(inOrder[j])] = map[idx(preOrder[i])] - 1; //right subtree for (int j = map[idx(preOrder[i])] + 1; j <= subb[idx(preOrder[i])]; j ++){ suba[idx(inOrder[j])] = map[idx(preOrder[i])]; subb[idx(inOrder[j])] --; } //resort by post-order for (int j = map[idx(preOrder[i])] + 1; j <= subb[idx(preOrder[i])]; j ++){ inOrder[j - 1] = inOrder[j]; map[idx(inOrder[j])] --; } inOrder[subb[idx(preOrder[i])]] = preOrder[i]; map[idx(preOrder[i])] = subb[idx(preOrder[i])]; } // printf("%s/n", inOrder); fclose(stdin); fclose(stdout); return 0; }