HDU 5444 Elven Postman(重建二叉树)

最新推荐文章于 2020-11-29 17:04:10 发布

Algobird

最新推荐文章于 2020-11-29 17:04:10 发布

阅读量838

点赞数

CC 4.0 BY-SA版权

分类专栏：图

本文链接：https://blog.youkuaiyun.com/u012762625/article/details/48656929

图专栏收录该内容

9 篇文章

订阅专栏

本文介绍了如何使用先序和中序序列建立二叉树，并记录父亲节点和方向。通过递归方法输出解，利用哈希表记录节点地址，避免后续查找。重点复习了二叉树重建的相关算法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;
		
	}