HDU 5444 Elven Postman(重建二叉树)

来源：互联网发布：网络学硬笔书法编辑：程序博客网时间：2024/05/17 05:54

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5444

用先序和中序建立二叉树，同时记录父亲节点和方向。

然后递归输出解就可以了，用hash记录每个节点的地址，之后不需要再查找。

主要是复习一下二叉树重建。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <queue>#include <set>#include <stack>using namespace std;const int maxn = 1005;struct node {//    int label;    node* l;    node* r;    node* father;    char dir;    node (int lb) { //       label = lb;        l = r = father = 0;        dir = ' ';    }    ~node() {        delete l;        delete r;    }};int preOrder[maxn];int n;node *hs[maxn]; // 用来映射树的节点的地址node* build(int id, int s, int e, node* father, char dir) {    if(s > e) {        return 0;    }    int root = preOrder[id];    node* cur = new node(root);    cur->father = father;    cur->dir = dir;    hs[root] = cur;    cur->l = build(id + 1, s, root - 1, cur, 'E');    cur->r = build(id + (root - s) + 1, root + 1, e, cur, 'W');    return cur;}//void preOrder1(node* root) {//    if(root == 0) {//        return ;//    }//    printf("%d", root->label);//    preOrder1(root->l);//    preOrder1(root->r);//}void print(node* cur) {    if(cur->dir == 'r') {        return ;    }    print(cur->father);    printf("%c", cur->dir);}int main() {   int t, q, x;//   freopen("test.txt", "r", stdin);   scanf("%d", &t);   while (t --) {       scanf("%d", &n);       for (int i = 1; i <= n; i ++) {            scanf("%d", preOrder + i);       }       node *root = build(1, 1, n, 0, 'r');//       preOrder1(root);       scanf("%d", &q);       while (q --) {            scanf("%d", &x);            node *cur = hs[x]; //           printf("%d %d\n", cur->father, cur->dir);            print(cur);            printf("\n");       }   }   return 0;}

附上一个以前写的二叉树重建：方便自己以后回忆。

// 根据先序序列和中序序列，恢复二叉树public BinNode<T> reBuilderPre(Object[] pre, Object[] ino) {root = reBuilderPre(pre, ino, 0, 0, ino.length - 1);return root;}private BinNode<T> reBuilderPre(Object[] pre, Object[] ino, int ps, int low, int high) {if(low > high) {return null;}T elem = (T) pre[ps];BinNode<T> p = new BinNode<T>((T) pre[ps]);int i = 0;for (i = low; i <= high; i ++) {if(elem.equals(ino[i])) {break;}}p.lchild = reBuilderPre(pre, ino, ps + 1, low, i - 1);p.rchild = reBuilderPre(pre, ino, ps + (i - low) + 1, i + 1, high);return p;}// 根据后序序列和中序序列，恢复二叉树public BinNode<T> reBuilderPost(Object[] post, Object[] ino) {root = reBuilderPost(post, ino, post.length - 1,  0, ino.length - 1);return root;}private BinNode<T> reBuilderPost(Object[] post, Object[] ino, int ps,  int low, int high) {if(low  > high) {return null;}T elem = (T)post[ps];BinNode<T> p = new BinNode<T>((T)post[ps]);int i = 0;for (i = low; i < high; i ++) {if(elem.equals(ino[i])) {break;}}p.lchild = reBuilderPost(post, ino, ps - (high - i) - 1, low, i - 1);p.rchild = reBuilderPost(post, ino, ps - 1, i + 1, high);return p;}

0 0