#642 (Div. 3)D. Constructing the Array（优先队列）

题目描述

You are given an array a of length n consisting of zeros. You perform n actions with this array: during the i-th action, the following sequence of operations appears:
Choose the maximum by length subarray (continuous subsegment) consisting only of zeros, among all such segments choose the leftmost one;
Let this segment be [l;r]. If r−l+1 is odd (not divisible by 2) then assign (set) a[l+r2]:=i (where i is the number of the current action), otherwise (if r−l+1 is even) assign (set) a[l+r−12]:=i.
Consider the array a of length 5 (initially a=[0,0,0,0,0]). Then it changes as follows:
Firstly, we choose the segment [1;5] and assign a[3]:=1, so a becomes [0,0,1,0,0];
then we choose the segment [1;2] and assign a[1]:=2, so a becomes [2,0,1,0,0];
then we choose the segment [4;5] and assign a[4]:=3, so a becomes [2,0,1,3,0];
then we choose the segment [2;2] and assign a[2]:=4, so a becomes [2,4,1,3,0];
and at last we choose the segment [5;5] and assign a[5]:=5, so a becomes [2,4,1,3,5].
Your task is to find the array a of length n after performing all n actions. Note that the answer exists and unique.
You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (1≤n≤2⋅105) — the length of a.
It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 (∑n≤2⋅105).

Output

For each test case, print the answer — the array a of length n after performing n actions described in the problem statement. Note that the answer exists and unique.

Example

input
6
1
2
3
4
5
6
output
1
1 2
2 1 3
3 1 2 4
2 4 1 3 5
3 4 1 5 2 6

题目大意

给你一个全为0的数组 要求把 1 到 n 插入数字，插入规则为：第i次选择一个左边最长的全0区间，将区间 (l+r)/2的位置变为 i。

题目分析

用优先队列维护一个区间集合。每次取出长度最大的区间。
将中点赋值，再从中点将这个区间分割为成两个，再放入队列中。
这样求出每个点的值来。

代码如下

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include  
#include 
#include 
#define LL long long
using namespace std;
int const N=2e5+5;
struct Node{
	int l,r;
	bool operator< (const Node a) const   //定义比较准则
	{
		if(r-l==a.r-a.l) return l>a.l;   //长度相等时，找靠前的
		return r-l<a.r-a.l;              //否则找长度大的
	}
};
int ans[N];     //答案数组
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		memset(ans,0,sizeof ans);    //清空数组
		priority_queue<Node> heap;
		heap.push({1,n});
		
		int i=1;      //第i次操作
		while(heap.size())
		{
			Node t=heap.top();
			heap.pop();
			
			int len=t.r-t.l+1;
			int k=(t.l+t.r)/2;
			
			ans[k]=i;       //将中点进行赋值
			if(len==2) heap.push({t.r,t.r}); //长度为2，只需要放入后面的那一个点
			else if(len>2)       //长度大于2，则将区间分成两段
			{
			    heap.push({t.l,k-1});
			    heap.push({k+1,t.r});
		    }
			i++;
		}
		for(int i=1;i<=n;i++)     //输出答案序列
		printf("%d ",ans[i]);
		puts("");
	}
	return 0;
}