POJ2828 Buy Tickets（线段树)

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

40 771 511 332 6940 205231 192431 38900 31492

Sample Output

77 33 69 5131492 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：有一个买票队列，给出每个人的插队的序号和值，根据插队的顺序输出最后的队列的值 比如0 771 511 332 69首先 0 77 则第一个位置是77 然后1 51 说明这个人插在第一个人的后面 此时第二个位置是51 然后是1 33 说明这个人又插在第一个人后面 此时第二个位置是33 第三个位置由原来第二个位置的人被挤过来 是51 最后2 69 说明插在第二个人后面说明此时第三个位置是69 原来第三个位置又被挤到第四位 最后的序列是77 33 69 51思路：比较神的一个题...各种请教学长 各种百度题解 要从后往前插 就拿上面的数据来说吧 最后一个是2 69 说明这个人是插在第三个位置 他的前面还有2个人 因为他是最后一个插入的 后面没有人 所以他排名第三的位置是不可动摇的 因为是从后往前插 此时他前面还没有人 需要保留2个空位给以后的2个人插 然后1 33 说明这人前面有一个人 需要给那个人保留一个空位 所以他可以放在第二个位置 他的位置也是不可动摇的了 接着1 51 他也需要给前面保留一个空位 但是第二位、第三位已经有人了 所以他只能在第四位 最后一个0 77放在第一位线段树用来保存这个区间有多少个空位***********************************************************************************************************************/#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <stack>#include <vector>#include <ctime>using namespace std;#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1#define LEN 222222int sum[LEN << 2] , pos[LEN] , val[LEN] , ans[LEN];void PushUp(int rt){sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void build(int l , int r , int rt){if(l == r){sum[rt] = 1;return ;}int m = (l + r) >> 1;build(lson);build(rson);PushUp(rt);}int update(int p , int l , int r , int rt){if(l == r){sum[rt] = 0;return l;}int m = (l + r) >> 1 , ret;if(sum[rt << 1] >= p) ret = update(p , lson);else{p -= sum[rt << 1];ret = update(p , rson);}PushUp(rt);return ret;}int main(){//freopen("data.in" , "r" , stdin);int n;while(~scanf("%d", &n)){bool first = true;build(0 , n - 1 , 1);for(int i = 0 ; i < n ; i ++)scanf("%d%d", &pos[i] , &val[i]);for(int i = n - 1 ; i >= 0 ; i --){int p = update(pos[i] + 1, 0 , n - 1 , 1);ans[p] = val[i];}for(int i = 0 ; i < n ; i ++){if(first){printf("%d" , ans[i]);first = false;}elseprintf(" %d", ans[i]);}puts("");}return 0;}