HDU 5444 Elven Postman 二叉排序树

HDU 5444

题意：给你一棵树的先序遍历，中序遍历默认是1...n，然后q个查询，问根节点到该点的路径（题意挺难懂，还是我太傻逼）

思路:这他妈又是个大水题，可是我还是太傻逼。1000个点的树，居然用标准二叉树结构来存点，，，卧槽想些什么东西。可以用一维数组，left,right直接指向位置就行了，我这里用的是链表。对于读入的序列，生成一个二叉排序树，同时记录一下路径就可以了，后面居然忘了清空path又wa了一发。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<string>
#define inf 0x3f3f3f3f
#define LL long long
#define MAXN 2005
#define IN freopen("in.txt","r",stdin);
using namespace std;

struct Node{
    int x;
    Node *left, *right;
    Node(int x = 0) :x(x){
        left = NULL;
        right = NULL;
    };
};

vector<char> path[MAXN];

Node *root;
void bst_insert(int x){
    //Node *t = root;
    //return;
    Node *t = root;
    //Node *s = &Node(x);
    while (t != NULL){
        if (x > t->x){
            path[x].push_back('W');
            if (t->right == NULL){
                t->right = new Node(x);
                break;
            }
            else{
                t = t->right;
            }
        }
        else{
            path[x].push_back('E');
            if (t->left == NULL){
                t->left = new Node(x);
                break;
            }
            else{
                t = t->left;
            }
        }
    }
}

int main(){
    //IN;
    //freopen("out.txt", "w", stdout);
    int t, n;
    scanf("%d", &t);
    while (t--){
        scanf("%d", &n);
        int x;
        scanf("%d", &x);
        root = new Node(x);
        for (int i = 1; i <= n; i++){
            path[i].clear();
        }
        for (int i = 2; i <= n; i++) {
            scanf("%d", &x);
            bst_insert(x);
        }
        //    find(1, n, 1);
        int q;
        scanf("%d", &q);
        //bfs();
        while (q--){
            scanf("%d", &x);
            for (int i = 0; i < path[x].size(); i++){
                printf("%c", path[x][i]);
            }
            printf("
");
        }
    }
    return 0;
}