poj-2828 Buy Tickets(经典线段树)

/*
Buy Tickets
Time Limit: 4000MS        Memory Limit: 65536K
Total Submissions: 10207        Accepted: 4919
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 Posi and Vali in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Posi and Vali are as follows:

Posi ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Posi-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.
Vali ∈ [0, 32767] — The i-th person was assigned the value Vali.
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.

Source

POJ Monthly--2006.05.28, Zhu, Zeyuan

//一般题意翻译我都是摘抄的，呵呵
排队买票，依次给出当前人要插队的位置，然后问你最后整个的序列是神马？

这个题目很想往线段树上考虑，后来看了题解，大牛说的是这样，由于左后一个人插进来后他的位置肯定是固定的

我们就可以倒着来插，最后一个固定后，如果倒数第二个插入的序号小于当前那么就往前插到序号上，否则往后插，往后的话序号需要减去当前这个数左边的空位数

因为左右都是从0位置开始标记的

因此结构体里需要维持节点左右边的空位个数，当前插队序号小于左边空位插左边，大于的话插右边，但是需要需要减去左边空位。。因为右边也是从0位置开始算起的，并且

我们是倒着插，所以空位个数才是当时的需要插的位置，已经被占的位置当时还不存在..
*/
//用时1454MS，嘿嘿比较长
#include <iostream>
#include<algorithm>
#include<queue>
#include<cmath>//math
#include<string.h>//memset(a,0,sizeof(a));
#include<stdio.h>
#include<stdlib.h>//system("pause");
using namespace std;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
const int maxn=;
int sum[maxn<<];
int x[maxn];
struct Node
{
    int p,sr;
}a[maxn];
void PushUp(int rt)
{
    sum[rt]=sum[rt<<]+sum[rt<<|];
}
void cre(int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]=;
        return ;
    }
    int m=(l+r)>>;
    cre(lson);
    cre(rson);
    PushUp(rt);
}
void upd(int p,int sr,int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]=;
        x[l]=sr;
        //printf("x[%d] = %d\n",l,x[l]);
        return ;
    }
    int m = (l + r)>>;
    if(p<sum[rt<<])
        upd(p,sr,lson);
    else
        upd(p-sum[rt<<],sr,rson);
    PushUp(rt);
}
int main()
{
    int n,i;
    while(~scanf("%d",&n))
    {
        cre(,n-,);
        for(i=;i<n;i++)
        scanf("%d%d",&a[i].p,&a[i].sr);
        cre(,n-,);
        for(i=n-;i>=;i--)
        {
            upd(a[i].p,a[i].sr,,n-,);
        }
        for(i=;i<n-;i++)
        {
            printf("%d ",x[i]);
        }
        printf("%d\n",x[i]);
    }
    return ;
}