zoukankan      html  css  js  c++  java
  • Codeforces Round #353 (Div. 2) D. Tree Construction 模拟

    D. Tree Construction

    题目连接:

    http://www.codeforces.com/contest/675/problem/D

    Description

    During the programming classes Vasya was assigned a difficult problem. However, he doesn't know how to code and was unable to find the solution in the Internet, so he asks you to help.

    You are given a sequence a, consisting of n distinct integers, that is used to construct the binary search tree. Below is the formal description of the construction process.

    First element a1 becomes the root of the tree.
    Elements a2, a3, ..., an are added one by one. To add element ai one needs to traverse the tree starting from the root and using the following rules:
    The pointer to the current node is set to the root.
    If ai is greater than the value in the current node, then its right child becomes the current node. Otherwise, the left child of the current node becomes the new current node.
    If at some point there is no required child, the new node is created, it is assigned value ai and becomes the corresponding child of the current node.

    Input

    The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the length of the sequence a.

    The second line contains n distinct integers ai (1 ≤ ai ≤ 109) — the sequence a itself.

    Output

    Output n - 1 integers. For all i > 1 print the value written in the node that is the parent of the node with value ai in it.

    Sample Input

    3
    1 2 3

    Sample Output

    1 2

    Hint

    题意

    给你平衡树的节点,每个节点都依次插进去的,问你每个节点插进去之后,他的父亲是谁

    题解:

    没必要真的写个平衡树,直接用stl模拟就好了……

    代码

    #include<bits/stdc++.h>
    using namespace std;
    set<int> s;
    map<int,int> ls,rs;
    int x;
    int main()
    {
        int n;
        scanf("%d",&n);
        scanf("%d",&x);
        s.insert(x);
        for(int i=1;i<n;i++)
        {
            scanf("%d",&x);
            auto it=s.lower_bound(x);
            if(it==s.end())
            {
                it--;
                rs[*it]=x;
                printf("%d ",*it);
            }
            else if(ls[*it]==0)
            {
                ls[*it]=x;
                printf("%d ",*it);
            }
            else
            {
                it--;
                rs[*it]=x;
                printf("%d ",*it);
            }
            s.insert(x);
        }
        printf("
    ");
    }
  • 相关阅读:
    zabbix执行远程命令
    zabbix监控主从同步
    s3fs+minio模拟挂载S3服务器到本地服务器
    PHP编译报错
    ldd可执行程序时返回not a dynamic executable
    Windows nessus安装
    Django数据库,在原有表中添加新字段
    docker安装fastdfs与java客户端测试
    Docker安装与启动
    2018HBCPC
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5503453.html
Copyright © 2011-2022 走看看