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[30];
int root,lson[30],rson[30],fa[30];
string str;
int dfs(string s,int fa) {
    if(s.length()<1) return 0;
    int u=s[0]-'A'+1;
    if(s.length()==1) {return u;}
    lson[u]=dfs(s.substr(1,id[u]-id[fa]-1),fa);
    rson[u]=dfs(s.substr(id[u]-id[fa]),u);
    return u;
}
void ans(int u) {
    if(u==0) return;
    ans(lson[u]);
    ans(rson[u]);
    printf("%c",u+'A'-1);
}
void init() {
    cin>>str;
    xiaosiji(i,0,str.length()) {
        id[str[i]-'A'+1]=i+1;    
    }
    cin>>str;
    root=str[0]-'A'+1;
}
void solve() {
    init();
    dfs(str,0);
    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();
}