HDU 5444 Elven Postman 二叉排序树

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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 0x3f3f3f3f11 #define LL long long12 #define MAXN 200513 #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 }

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