【平衡二叉树】Buy Tickets

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.

看到这道题，第一个想到的是跳跃表，但是超时超空间。
于是改用SBT，很快就过了。
插入的时候，并不按关键字检索，而是根据位置插入。
Accode：

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <string>

const int maxN = 400010, INF = 0x3f3f3f3f;
int sz[maxN], lc[maxN];
int rc[maxN], key[maxN];
int n, T, tot;

inline void Zig(int &T)
{
    int tmp = lc[T]; lc[T] = rc[tmp];
    rc[tmp] = T; sz[tmp] = sz[T];
    sz[T] = sz[lc[T]] + sz[rc[T]] + 1;
    T = tmp; return;
}

inline void Zag(int &T)
{
    int tmp = rc[T]; rc[T] = lc[tmp];
    lc[tmp] = T; sz[tmp] = sz[T];
    sz[T] = sz[lc[T]] + sz[rc[T]] + 1;
    T = tmp; return;
}

void maintain(int &T)
{
    if (sz[lc[lc[T]]] > sz[rc[T]]) Zig(T);
    else if (sz[rc[lc[T]]] > sz[rc[T]])
    {Zag(lc[T]); Zig(T);}
    else if (sz[rc[rc[T]]] > sz[lc[T]]) Zag(T);
    else if (sz[lc[rc[T]]] > sz[lc[T]])
    {Zig(rc[T]); Zag(T);}
    else return;
    maintain(lc[T]); maintain(rc[T]);
    maintain(T); return;
}

void Ins(int &T, int v, int k)
{
    if (!T)
    {
        sz[T = ++tot] = 1;
        key[tot] = v;
        return;
    }
    ++sz[T];
    if (k <= sz[lc[T]])
        Ins(lc[T], v, k);
    else Ins(rc[T], v, k - sz[lc[T]] - 1);
    maintain(T); return;
}

void Inorder(int T)
{
    if (lc[T]) Inorder(lc[T]);
    if (key[T] < INF) printf("%d ", key[T]);
    if (rc[T]) Inorder(rc[T]);
    return;
}

inline int getint()
{
    int res = 0; char tmp;
    while (!isdigit(tmp = getchar()));
    do res = (res << 3) + (res << 1) + tmp - '0';
    while (isdigit(tmp = getchar()));
    return res;
}

int main()
{
    freopen("ticket.in", "r", stdin);
    freopen("ticket.out", "w", stdout);
    while (scanf("%d", &n) != EOF)
    {
        memset(key, 0, sizeof key);
        memset(sz, 0, sizeof sz);
        memset(lc, 0, sizeof lc);
        memset(rc, 0, sizeof rc);
        T = tot = 0; //注意清零。
        for (int i = 0; i < n; ++i)
        {
            int pos = getint(), val = getint();
            Ins(T, val, pos);
        }
        Inorder(T);
        printf("\n");
    }
    return 0;
}