POJ - 2828 Buy Tickets —— 插空

最新推荐文章于 2018-09-05 16:00:25 发布

原创最新推荐文章于 2018-09-05 16:00:25 发布 · 338 阅读

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探讨了在特定条件下人们加入队列并插队时如何确定最终队列顺序的问题，利用线段树进行高效的数据结构维护。

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Buy Tickets

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 21897		Accepted: 10734

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_iin 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：共有n个人，每个人的价值为v，来了以后直接插在p的位置上，其后的人向后退一位，问所有人插完最后的序列

思路：从后向前遍历插入，每个人插入的位置是从队首开始有p个空位的地方，比如说第二组样例中，31492的p为0，直接插在队首下标0的位置，3890的p为1，需要从队首数出一个空位（下标为1），插在下标为2的位置，依此类推

原因：每次插入时，插入位置如果为p，则其前一定已存在p个数

从队首开始数的时候，有数字的位置表示这个数是在当前插入数的后来插入的，那要保证插入成功，就得在当前数前面留下p个空格等待插入

用线段树来维护各个区间中空格的个数

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <cstring>
#include <queue>
#include <algorithm>
#define ll long long
#define max_ 201000
#define inf 0x3f3f3f3f
#define mod 1000000007
using namespace std;
struct node
{
	int l,r,w;
	int sum;
};
struct node tree[max_*4];
int n;
int num[max_],pos[max_];
void built(int i,int l,int r)
{
	tree[i].l=l;
	tree[i].r=r;
	if(l==r)
	{
		tree[i].sum=1;
		return;
	}
	int mid=(l+r)>>1;
	built(i<<1,l,mid);
	built(i<<1|1,mid+1,r);
	tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;
}
void updata(int i,int cnt,int v)
{
	if(tree[i].l==tree[i].r)
	{
		tree[i].w=v;
		tree[i].sum--;
		return;
	}
	if(tree[i<<1].sum>cnt)
	updata(i<<1,cnt,v);
	else
	updata(i<<1|1,cnt-tree[i<<1].sum,v);
	tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;
}
void show(int i)
{
	if(tree[i].l==tree[i].r)
	{
		printf("%d ",tree[i].w);
		return;
	}
	show(i<<1);
	show(i<<1|1);
}
int main(int argc, char const *argv[]) {
	while(scanf("%d",&n)!=EOF)
	{
		built(1,1,n);
		for(int i=1;i<=n;i++)
			scanf("%d%d",&pos[i],&num[i]);
		for(int i=n;i>=1;i--)
			updata(1,pos[i],num[i]);
		show(1);
		printf("\n" );
	}
	return 0;
}