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;
}