POJ 2828 Buy Tickets（线段树）

题意：火车站有n个人排队，他们是按顺序到达的，但是他们乱插队。每个人有两个值pos[i]和val[i]。比如现在第5个人来了，他的pos[5]值为3，那么他就会插队到当前第3个人位置的后面(第0个人是售票窗口)。依次给出所有人的pos和val值，要你最终按所有人的位置顺序输出val值。

思路：维护一颗线段树，该树的叶子结点初始值为1，表示这个位置可以占用，当确定一个人的位置的时候就把那个值变成0.显然如果从前往后推的话很困难，每下一个人的插入又会改变前面的人的位置，所以呢我们从后往前插入，比如最后一个人的位置就肯定是固定的了，那么固定完这个位置之后不管他，倒数第二个人的位置也一定是固定的了。

Trick:用G++提交超时了...用C++过了

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 200000+100;
#define lson i*2,l,m
#define rson i*2+1,m+1,r

struct Node
{
	int p,v;
}node[maxn];
int val[maxn*4];
int sum[maxn*4];
int cnt,n;

void PushUp(int i)
{
	sum[i] = sum[i*2]+sum[i*2+1];
}

void build(int i,int l,int r)
{
	if (l==r)
	{
		sum[i]=1;
		return;
	}
	int m = (l+r)/2;
	build(lson);
	build(rson);
	PushUp(i);
}

void update(int p,int v,int i,int l,int r)
{
	if (l==r)
	{
		if (sum[i])
		{
			sum[i]=0;
			val[i]=v;
		}
		return;
	}
	int m = (l+r)/2;
	if (p<=sum[i*2])
		update(p,v,lson);
	else
		update(p-sum[i*2],v,rson);
	PushUp(i);
}

void print(int i,int l,int r)
{
    if (l==r)
	{
		cnt++;
		printf("%d",val[i]);
		if (cnt<n)
			printf(" ");
		else
			printf("\n");
		return;
	}
	int m = (l+r)/2;
	print(lson);
	print(rson);
}

int main()
{
	while (scanf("%d",&n)!=EOF && n)
	{
		cnt = 0;
		build(1,1,n);
		for (int i = 1;i<=n;i++)
		{
			scanf("%d%d",&node[i].p,&node[i].v);
			node[i].p++;
		}
		for (int i = n;i>=1;i--)
		{
			update(node[i].p,node[i].v,1,1,n);
		}
		print(1,1,n);
	}
}

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_iand 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

4
0 77
1 51
1 33
2 69
4
0 20523
1 19243
1 3890
0 31492

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.

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