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(" "); 84 } 85 } 86 return 0; 87 }