codeforces 675D 模拟构建二叉排序树

迈不过友情╰ 2022-07-24 07:23 126阅读 0赞

D. Tree Construction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

1.  First element *a*1 becomes the root of the tree.
2.  Elements *a*2, *a*3, ..., *a**n* are added one by one. To add element *a**i* one needs to traverse the tree starting from the root and using the following rules:
    
    1.  The pointer to the current node is set to the root.
    2.  If *a**i* 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.
    3.  If at some point there is no required child, the new node is created, it is assigned value *a**i* 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 *a**i* (1 ≤ *a**i* ≤ 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 *a**i* in it.

Examples

input

3
    1 2 3

output

1 2

input

5
    4 2 3 1 6

output

4 2 2 4

Note

Picture below represents the tree obtained in the first sample.

![f2f3bf8a19408d1f82699a41b787e56a6452dd81.png][]

Picture below represents the tree obtained in the second sample.

![f5bdbb65dc9bd752e030a47d4808c15b1b19aae1.png][]

题意：把每个数放进二叉排序树中，输出从第二个数到第n个数的直接父节点。

分析：刚开始写了一个正规的排序二叉树，然后不断插入数据，插入数据的过程中同时记录这个数的父节点，但是没考虑到数据量太大，而且排序二叉树可能会退化成链表的模样，使得插入的时间复杂度变成O（n^2）级别，很明显会超时。后来看题解，用一个set和两个map来模拟二叉树，查找的时候用二分查找， 这样使得复杂度就在O（nlogn）级别了。

先贴一下超时的代码：

#include<bits/stdc++.h>
    
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;int f=1;
        while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res*f;
    }
    const int N=100010,MOD=1e9+7;
    int a[N];
    
    struct node
    {
        int val;
        node *l,*r;
    };
    int ret;
    node *insert(node *p,int x)
    {
        if(p==NULL)
        {
            node *q = new node;
            q->val=x;
            q->l=q->r=NULL;
            return q;
        }
        else
        {
            if(x <= p->val)
            {
                ret = p->val;
                p->l=insert(p->l,x);
            }
            else {
                    ret = p->val;
                    p->r=insert(p->r,x);
            }
            return p;
        }
    }
    
    int find(node *p,int x,int fa)
    {
        if(x==p->val) return fa;
        else if(x <= p->val)
        {
            return find(p->l,x,p->val);
        }
        else
        {
            return find(p->r,x,p->val);
        }
    }
    
    int main()
    {
        int n=in();
        node * root = new node;
        root->l = root->r = NULL;
        root->val = in();
        for(int i=1;i<n;i++)
        {
            a[i]=in();
            ret = a[0];
            root=insert(root,a[i]);
            printf("%d ",ret);
        }
    //    for(int i=1;i<n;i++){
    //        printf("%d ",find(root,a[i],a[0]));
    //    }
        return 0;
    }

AC代码：

#include<bits/stdc++.h>
    
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;int f=1;
        while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res*f;
    }
    const int N=100010,MOD=1e9+7;
    map<int,int> Left,Right;
    set<int> numbers;
    set<int>::iterator it;
    
    int main()
    {
        int n=in(),x=in(),ret;
        numbers.insert(x);
    
        for(int i=1;i<n;i++){
            x=in(); //数都是不同的
            it = numbers.upper_bound(x);
            //如果找到了第一个比x大的,并且这个点的左子树不存在
            if(it != numbers.end() && Left.count(*it) == 0){
                Left[*it] = x; //插入左边
                ret = *it;
            }
            //没找到或者找到了, 但是左子树存在,那么找到小于x的最大的点
            //(这个点应该在存在的左子树中),x应该是他的右孩子(排序二叉树的特性)
            else{
                it--;
                Right[*it] = x;
                ret = *it;
            }
            numbers.insert(x);
            printf("%d ",ret);
        }
        return 0;
    }

[f2f3bf8a19408d1f82699a41b787e56a6452dd81.png]: /images/20220722/10f2918c89e443b8b57d9dd4b1431d8f.png
[f5bdbb65dc9bd752e030a47d4808c15b1b19aae1.png]: /images/20220722/2f6a2541a6a443129f56fc4d7f30ac6e.png