usaco American Heritage

正常的想法是先通过先序和中序串构造一棵二叉树，然后后序遍历即可

对于先序中的结点x，先通过中序串找到其在中序中的位置i，如果该节点有左子树则vis[i - 1] = true，如果该节点的有右子树则vis[i + 1] = true

！！！传递的参数要用引用类型

/*
ID: daijinq1
PROB:heritage
LANG: C++
*/
#include <iostream>
#include <sstream>
#include <string>
#include <cstdio>
using namespace std;

#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define FORD(i,a,b) for(i = (a); i > (b); --i)
#define FORDE(i,a,b) for(i = (a); i >= (b); --i)
#define CLR(a,b) memset(a,b,sizeof(a))

const int MAXN = 10000;
const int MAXM = 0;
const int hash_size = 25000002;
const int INF = 0x7f7f7f7f;

bool vis[30];
string mid_root, fir_root;
struct node {
char ch;
node* left;
node* right;
};

node* root;

void end_root_scan(node* temp_root)
{
if(temp_root == NULL)
return ;
end_root_scan(temp_root->left);
end_root_scan(temp_root->right);
cout<<temp_root->ch;
}

void create(int &temp_root, node* &root_tree) {
// 指针引用
int i;

root_tree = new node;
root_tree->ch = fir_root[temp_root];
root_tree->left = NULL;
root_tree->right = NULL;
bool left_sign = false, right_sign = false;

for(i = 0; i < (int)mid_root.length(); ++i)
if(mid_root[i] == fir_root[temp_root]) {
vis[i] = true;
break;
}
if(i > 0 && !vis[i - 1] && temp_root < (int)mid_root.length() - 1)
create(++temp_root, root_tree->left);
if(i < (int)mid_root.length() - 1 && !vis[i + 1] && temp_root < (int)mid_root.length() - 1)
create(++temp_root, root_tree->right);
return ;
}

void input() {
cin>>mid_root>>fir_root;
int x = 0;
create(x, root);
end_root_scan(root);
cout<<endl;
}

int main() {
freopen("heritage.in", "r", stdin);
freopen("heritage.out", "w", stdout);
input();
return 0;
}