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。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<queue>
 7 #include<vector>
 8 #include<set>
 9 #include<string>
10 #define inf 0x3f3f3f3f
11 #define LL long long
12 #define MAXN 2005
13 #define IN freopen("in.txt","r",stdin);
14 using namespace std;
15
16 struct Node{
17     int x;
18     Node *left, *right;
19     Node(int x = 0) :x(x){
20         left = NULL;
21         right = NULL;
22     };
23 };
24
25 vector<char> path[MAXN];
26
27 Node *root;
28 void bst_insert(int x){
29     //Node *t = root;
30     //return;
31     Node *t = root;
32     //Node *s = &Node(x);
33     while (t != NULL){
34         if (x > t->x){
35             path[x].push_back(‘W‘);
36             if (t->right == NULL){
37                 t->right = new Node(x);
38                 break;
39             }
40             else{
41                 t = t->right;
42             }
43         }
44         else{
45             path[x].push_back(‘E‘);
46             if (t->left == NULL){
47                 t->left = new Node(x);
48                 break;
49             }
50             else{
51                 t = t->left;
52             }
53         }
54     }
55 }
56
57 int main(){
58     //IN;
59     //freopen("out.txt", "w", stdout);
60     int t, n;
61     scanf("%d", &t);
62     while (t--){
63         scanf("%d", &n);
64         int x;
65         scanf("%d", &x);
66         root = new Node(x);
67         for (int i = 1; i <= n; i++){
68             path[i].clear();
69         }
70         for (int i = 2; i <= n; i++) {
71             scanf("%d", &x);
72             bst_insert(x);
73         }
74         //    find(1, n, 1);
75         int q;
76         scanf("%d", &q);
77         //bfs();
78         while (q--){
79             scanf("%d", &x);
80             for (int i = 0; i < path[x].size(); i++){
81                 printf("%c", path[x][i]);
82             }
83             printf("\n");
84         }
85     }
86     return 0;
87 }