POJ-2828 Buy Tickets【线段树 单点更新】

POJ-2828 Buy Tickets【线段树 单点更新】

Time Limit: 4000MS Memory Limit: 65536K

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.

题意：n个人顺序排队，每个人的信息为（posi，Vali），即这个要入队的站在第二个人的后面而有value值，结果输出队的顺序【value值】，如案例
题解：本来想用链表来做的数据量10^8太大，看大神做的都是用线段树来做，首先输入更新顺序和输出的顺序相反，把数组反向，

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>using namespace std;#define lid (id<<1) //左子树#define rid (id<<1|1)//id*2+1 右子树typedef long long LL;int n, h, k;const int N = 2000005;struct Segtree{    int l,r;//左、右子树    int num;//位置}tr[N];int a[N],po[N],va[N];void push_up(int id){//求和——位置    tr[id].num = tr[lid].num + tr[rid].num;}void build(int id,int l,int r){//建树    tr[id].l = l;    tr[id].r = r;    if(l == r){//叶子节点        tr[id].num = 1;//只能放入一人         return;    }    int mid = (l+r)>>1;    build(lid, l, mid);    build(rid, mid+1, r);    push_up(id);}void updata(int id,int v,int cur){//更新    if(tr[id].l == tr[id].r){//找到叶子节点po[i]，          tr[id].num = 0;//已放入，人数清0        a[ tr[id].l ] = v;//保存该人value        return;    }    int mid = (tr[id].l + tr[id].r)>>1;    if(tr[lid].num < cur){//根据左右儿子空位的多少和插入数据的位置比较来确定插入哪个儿子        updata( rid, v, cur - tr[lid].num );    }else {        updata( lid, v, cur );    }    push_up(id);}int main(){    while(~scanf("%d",&n)){        build (1, 1, n);        for(int i=1;i<=n;i++) scanf("%d %d",&po[i],&va[i]);        for(int i=n;i>=1;i--) updata(1,va[i],po[i]+1);//反向        for(int i=1;i<=n;i++){            printf("%d%c",a[i],(i==n?'\n':' '));        }    }    return 0;}