poj 2828 Buy Tickets 萬能的線段樹大法。

Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 14400   Accepted: 7199

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

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.

 

 

 

我AC的狀態：，，感覺不太滿意，有點慢了，但是我比較懶，不想在改進了~~

說實話，這題從良心上說，弄懂了，真簡單。但是我的弱項一直是線段樹，所以就在拼命的刷線段樹題，可是感覺長進不大。。囧 。。

解題思路：

 

觀察發現，最後一個插入到該位置的人位置是固定的，那麼我們可以從後面進行插入操作，pos ，val 代表val要插入到pos位置，那麼就是說 pos 前面要留出 pos個位置，

因為 是從 0 開始的。

線段樹 ：st數組記錄 該區間 目前還剩 x個空位，每一次 query即插入的時候 ，如果 該節點左兒子 x>=pos，那麼只要在左兒子找就可以了

否則 要在右兒子中 找 ，此時 pos 改為 pos-左兒子空位。。

 

注意紅體大字，，一開始就是沒理解這句話，一直在死扣題目。如果弄懂了上面的兩句話，那簡直就是so easy啦！

下面代碼：

#include 

#define MAX 200100

int st[MAX<<4],data[MAX];

void build(int left, int right ,int pos)
{
	if(left == right)
	{
		st[pos] = 1;
		return ;
	}
	int mid = (left+right)>>1 ;
	build(left,mid,pos<<1) ;
	build(mid+1,right,pos<<1|1) ;
	st[pos] = st[pos<<1]+st[pos<<1|1] ;
}

void update(int d , int x , int L , int R , int pos)
{
	if(R == L)
	{
		data[L] = d ;
		st[pos] = 0 ;
		return ;
	}
	int mid = (L+R)>>1 ;
	if(x<=st[pos<<1])
	{
		update(d,x,L,mid,pos<<1);
	}
	else
	{
		update(d,x-st[pos<<1],mid+1,R,pos<<1|1) ;
	}
	st[pos] = st[pos<<1]+st[pos<<1|1] ;
}

int main()
{
	int n,pos[MAX],val[MAX];
	while(~scanf(%d,&n))
	{
		build(1,n,1);
		for(int i = 0 ;i < n ; ++i)
		{
			scanf(%d%d,&pos[i],&val[i]) ;
		}
		for(int i = n-1 ; i >=0 ; --i)
		{
			update(val[i] , pos[i]+1 , 1 , n , 1);
		}
		for(int i = 1 ; i <= n ; ++i)
		{
			printf(%d,data[i]);
			if(i != n)
			{
				printf( ) ;
			}
		}
		printf(
) ;
	}
	return 0 ;
}