USACO 3.4 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 

—————————————————————
用前序遍历中某个字母的位置可以得到它的左子树的长度
然后就可以了
存代码

 /*
 ID: ivorysi
 PROG: heritage
 LANG: C++
 */
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #include <set>
 #include <vector>
 #include <string.h>
 #define siji(i,x,y) for(int i=(x);i<=(y);++i)
 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j)
 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i)
 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j)
 #define inf 0x3f3f3f3f
 #define MAXN 400005
 #define ivorysi
 #define mo 97797977
 #define ha 974711
 #define ba 47
 #define fi first
 #define se second
 #define pii pair<int,int>
 using namespace std;
 typedef long long ll;
 int id[];
 int root,lson[],rson[],fa[];
 string str;
 int dfs(string s,int fa) {
     if(s.length()<) return ;
     int u=s[]-'A'+;
     if(s.length()==) {return u;}
     lson[u]=dfs(s.substr(,id[u]-id[fa]-),fa);
     rson[u]=dfs(s.substr(id[u]-id[fa]),u);
     return u;
 }
 void ans(int u) {
     if(u==) return;
     ans(lson[u]);
     ans(rson[u]);
     printf("%c",u+'A'-);
 }
 void init() {
     cin>>str;
     xiaosiji(i,,str.length()) {
         id[str[i]-'A'+]=i+;
     }
     cin>>str;
     root=str[]-'A'+;
 }
 void solve() {
     init();
     dfs(str,);
     ans(root);
     puts("");
 }
 int main(int argc, char const *argv[])
 {
 #ifdef ivorysi
     freopen("heritage.in","r",stdin);
     freopen("heritage.out","w",stdout);
 #else
     freopen("f1.in","r",stdin);
 #endif
     solve();
 }