CodeForces - 675D Tree Construction （set&数据结构）

CodeForces - 675D

Tree Construction

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %I64d & %I64u

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Status

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
ndistinct 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

Input

3
1 2 3

Output

1 2

Input

5
4 2 3 1 6

Output

4 2 2 4

Sample Output

Hint

Source
Codeforces Round #353 (Div. 2)
//题意：
给你n个节点，第一个节点作为根节点，接下来的如果节点值>他的父节点，就将他练到他的右边，否则练到左边，输出从2--n节点的父节点的值。

//思路：
set模拟

#include<stdio.h>
#include<string.h>
#include<set>
#include<map>
#include<algorithm>
#include<iostream>
using namespace std;
set<int>t;
set<int>::iterator it;
set<int>::iterator it1;
set<int>::iterator it2;
map<int,int>pos;
int main()
{
int n,x;
while(scanf("%d",&n)!=EOF)
{
t.clear();
int T=1;
scanf("%d",&x);
t.insert(0);t.insert(x);
pos[x]=T++;
while(T<=n)
{
scanf("%d",&x);
it=t.lower_bound(x);
it1=it--;it2=it;
if(pos[*it1]<pos[*it2])
printf("%d ",*it2);
else
printf("%d ",*it1);
t.insert(x);
pos[x]=T++;
}
}
return 0;
}