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最新推荐文章于 2020-03-20 14:33:40 发布

Luwhere

最新推荐文章于 2020-03-20 14:33:40 发布

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分类专栏：算法文章标签：线段树

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本文探讨了一个购票排队的问题，通过描述在春运期间购票的场景，提出了一个关于人们如何在队伍中插队并保持秩序的问题。该文介绍了一种使用线段树的数据结构来解决此问题的方法。

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Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 15423		Accepted: 7685

Description

Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…

The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.

It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!

People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.

Input

There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The next N lines contain the pairs of values Pos_i and Val_i in the increasing order of i(1 ≤ i ≤ N). For each i, the ranges and meanings of Pos_i and Val_i are as follows:

Pos_i ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Pos_i-th person in the queue. The booking office was considered the 0th person and the person at the front of the queue was considered the first person in the queue.
Val_i ∈ [0, 32767] — The i-th person was assigned the value Val_i.

There no blank lines between test cases. Proceed to the end of input.

Output

For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.

Sample Input

Sample Output

77 33 69 51
31492 20523 3890 19243

Hint

The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input.

题意：

有N组序列Pi,Vi。表示Vi这个人插入Pi这个位置，若是有人在这个位置上则被挤向下一个位置，依次向后挤。输出最后的排队顺序。

题解：

可以将数据逆序读入，那么Pi则表示Pi这个位置之前有几个空位。用线段树记录空位个数解决。

参考代码：

#include<stdio.h>
#define M 200010
struct Tree{
	int r,l,cnt;
}tree[M<<2];
int result[M],a[M],b[M];
void build(int l,int r,int t)
{
	tree[t].l=l;
	tree[t].r=r;
	tree[t].cnt=r-l+1;//表示当前区间有几个空位。
	if(l==r)
	    return ;
    int mid=(l+r)>>1;
    build(l,mid,t<<1);
    build(mid+1,r,t<<1|1);
}
void update(int t,int pos,int v)
{
	tree[t].cnt--;//每次一个点进树就会使当前区间的空位减少。
	if(tree[t].r==tree[t].l){
		result[tree[t].l]=v;
		return ;
	}
	if(tree[t<<1].cnt>pos)//如果左子树的空位小于等于所需空位，那么去右子树。
	    update(t<<1,pos,v);
    else 
        update(t<<1|1,pos-tree[t<<1].cnt,v);//pos-tree[t<<1].cnt表示减去左边的空位数。
    
}
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		build(1,n,1);
		for(int i=1;i<=n;i++)
		{
		    scanf("%d%d",&a[i],&b[i]);
		}
		for(int i=n;i>=1;i--)
		{

			update(1,a[i],b[i]);
		}
		for(int i=1;i<=n-1;i++)
		    printf("%d ",result[i]);
        printf("%d\n",result[n]);
	}
	return 0;
}