线段树

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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.

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题目大意

有 n 个人排队买票，他们依次到来，第 i 个人来的时候会站在第pos[i]个人后面，并且他的编号为val[i]。

求最后的队列中每个位置人的编号。

题解

这题和poj2182可以说是一模一样。也是开一个线段树，维护当前区间中有几个空位，然后从后往前更新，在第几个空位插入这个人，并把该空位删除。可以自己手动模拟一下，具体可以参考。

代码

#include 
    
     using namespace std;#define pushup(u) {sum[u] = sum[u<<1] + sum[u<<1|1];}#define ls u<<1,l,mid#define rs u<<1|1,mid+1,rconst int maxn = 2e5 + 5;int sum[maxn << 2];int ans[maxn];int pos[maxn], id[maxn];void build(int u,int l,int r) {    sum[u] = r - l + 1;    if(l == r)return;    int mid = (l + r) >> 1;    build(ls);    build(rs);}void update(int u,int l,int r,int x,int a) {    if(l == r){        sum[u] = 0;        ans[l] = a;        return;    }    int mid = (l + r) >> 1;    if(sum[u << 1] >= x)update(ls,x,a);    else update(rs,x - sum[u<<1],a);    pushup(u);}int main() {    ios::sync_with_stdio(false); cin.tie(0);    int n;    while(cin >> n) {        build(1,1,n);        for(int i = 1;i <= n;i++) {            cin >> pos[i] >> id[i];        }        for(int i = n;i > 0;i--) {            update(1,1,n,pos[i] + 1,id[i]);        }        for(int i = 1;i <= n;i++) {            cout << ans[i];            i == n ? (cout << endl) : (cout << ' ');        }    }        return 0;}