排序二叉树 HDOJ 5444 Elven Postman

题目传送门

题意：给出线性排列的树，第一个数字是根节点，后面的数如果当前点小或相等往左走，否则往右走，查询一些点走的路径

分析：题意略晦涩，其实就是排序二叉树！1<<1000 普通数组开不下，用指针省内存。将树倒过来好理解些

E---------------------------------------W
                    2
                   / 
                  1   4
                      /
                     3

                     6
                    /
                   5
                  /
                 4
                / 
               3
              /
             2
            /
           1
E---------------------------------------W

收获：排序二叉树插入和查询

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/14 星期一 15:34:35
* File Name     :H.cpp
 ************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N], q[N];
struct BST {
    int val;
    BST *left, *right;
    BST *insert(BST *p, int x) {
        if (p == NULL)  {
            BST *t = new BST;
            t->val = x;
            t->left = t->right = NULL;
            return t;
        }
        if (x <= p->val)    p->left = insert (p->left, x);
        else    p->right = insert (p->right, x);
        return p;
    }
    bool find(BST *p, int x)   {
        if (x == p->val)    return true;
        else if (p == NULL) return false;
        else if (x <= p->val)    {
            cout << "E";
            return find (p->left, x);
        }
        else    {
            cout << "W";
            return find (p->right, x);
        }
    }
}bst;

int main(void)    {
    ios::sync_with_stdio (false);
    int T;  cin >> T;
    while (T--) {
        int n;  cin >> n;
        for (int i=1; i<=n; ++i)    {
            cin >> a[i];
        }
        int m;  cin >> m;
        for (int i=1; i<=m; ++i)    {
            cin >> q[i];
        }
        BST *root = NULL;
        for (int i=1; i<=n; ++i)    {
            root = bst.insert (root, a[i]);
        }
        for (int i=1; i<=m; ++i)    {
            bst.find (root, q[i]);
            cout << endl;
        }
    }

    return 0;
}