Elven Postman（根节点到指定节点的距离）

Elves are very peculiar creatures. As we all know, they can live for a very long time and their magical prowess are not something to be taken lightly. Also, they live on trees. However, there is something about them you may not know. Although delivering stuffs through magical teleportation is extremely convenient (much like emails). They still sometimes prefer other more “traditional” methods. 

So, as a elven postman, it is crucial to understand how to deliver the mail to the correct room of the tree. The elven tree always branches into no more than two paths upon intersection, either in the east direction or the west. It coincidentally looks awfully like a binary tree we human computer scientist know. Not only that, when numbering the rooms, they always number the room number from the east-most position to the west. For rooms in the east are usually more preferable and more expensive due to they having the privilege to see the sunrise, which matters a lot in elven culture. 

Anyways, the elves usually wrote down all the rooms in a sequence at the root of the tree so that the postman may know how to deliver the mail. The sequence is written as follows, it will go straight to visit the east-most room and write down every room it encountered along the way. After the first room is reached, it will then go to the next unvisited east-most room, writing down every unvisited room on the way as well until all rooms are visited. 

Your task is to determine how to reach a certain room given the sequence written on the root. 

For instance, the sequence 2, 1, 4, 3 would be written on the root of the following tree. 

Input

First you are given an integer T(T≤10)T(T≤10) indicating the number of test cases. 

For each test case, there is a number n(n≤1000)n(n≤1000) on a line representing the number of rooms in this tree. nn integers representing the sequence written at the root follow, respectively a1,...,ana1,...,an where a1,...,an∈{1,...,n}a1,...,an∈{1,...,n}. 

On the next line, there is a number qq representing the number of mails to be sent. After that, there will be qq integers x1,...,xqx1,...,xq indicating the destination room number of each mail.

Output

For each query, output a sequence of move (EE or WW) the postman needs to make to deliver the mail. For that EE means that the postman should move up the eastern branch and WW the western one. If the destination is on the root, just output a blank line would suffice. 

Note that for simplicity, we assume the postman always starts from the root regardless of the room he had just visited.

Sample Input

2
4
2 1 4 3
3
1 2 3
6
6 5 4 3 2 1
1
1

Sample Output

E

WE
EEEEE

题意：已知一个序列，建立一颗倒置的二叉搜索树。询问q次，每次查询二叉树上的某个值，输出从根节点到此点的路径。

思路：由于数据量略大，我们可以在建树的时候记录所有点的路径，在询问的时候直接输出即可。这个题要建的是一颗倒置的树，其实我们可以和平常一样建立顺序树，不过在输出路径的时候把路径倒置一下就好了。

代码如下：

#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<iostream>
#include<algorithm>
using namespace std;
string str;  //记录此时路径
map<int,string>v; //记录所有路径
struct node  //存储树的信息
{
    int val;
    node *lch,*rch;
};
node *insert(node *root,int x)  //建树
{
    if(root==NULL)   //此节点没有值
    {
        node *q=new node;  //声明
        q->val=x;      //节点赋值
        q->lch=q->rch=NULL;//此节点没有左右儿子
        return q;   //返回该节点指针
    }
    if(x<root->val)  //在左子树上
    {
        root->lch=insert(root->lch,x); //递归插入
        str+='E';  //记录路径
    }

    else if(x>root->val) //在右子树上
    {
         root->rch=insert(root->rch,x);//递归插入
         str+='W'; //记录路径
    }
    return root; //返回该节点指针
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        v.clear();
        scanf("%d",&n);
        node *root=NULL;
        str="";//路径为空
        for(int i=0;i<n;i++)
        {
            int x;
            string s=str;//记录一下上一次的路径，下次以此出发
            scanf("%d",&x);
            root=insert(root,x);
            v[x]=str;   //记录该节点的路径
            str=s;  //回到该路径上
        }
        scanf("%d",&n);
        while(n--)//查询
        {
            int x;
            scanf("%d",&x);
            string s=v[x];
            reverse(s.begin(),s.end());//路径倒置
            cout<<s<<endl;
        }
    }
}